let's solve this challenging derivative

problem step by step we'll see how

recognizing a special trigonometric

identity helps us simplify what looks

like a complex expression here's our

problem we need to find the derivative

of y = inverse tangent of the fraction

3x - x Cub over 1 -

3x^2 at first glance this looks quite

complicated however ever there's a key

Insight that will make this much easier

this fraction inside the inverse tangent

looks very similar to the triple angle

formula for tangent let's look at this

more closely if we let x equal tangent

Theta then our fraction becomes exactly

the triple angle formula for tangent

therefore our original expression can be

Rewritten as the inverse tangent of t

of 3 Theta and since inverse tangent and

tangent are inverse functions they

cancel out leaving us with just 3 Theta

now remember that x equals tangent Theta

so theta equals inverse tangent of X

therefore y = 3 * inverse tangent of X

now we can find the derivative the

derivative of inverse tangent is 1/ 1 +

x^2 and when we multiply by 3 we get our

final answer 3 over 1 + x^2 and that's

our solution what started as a

complicated looking derivative became

much simpler once we recognize the

triple angle formula for tangent this

shows how pattern recognition and

knowing your trigonometric identities

can greatly simplify calculus problems

Mastering Calculus: Derivative of Inverse Tangent Function [𝑡𝑎𝑛^(−1) ((3𝑥−𝑥^3)/(1−3𝑥^2 ))] Explained

Question
Find the derivative of the function y = [𝑡𝑎𝑛^(−1) ((3𝑥−𝑥^3)/(1−3𝑥^2 )]

In this video, we dive into the derivative of the inverse tangent function with a detailed, step-by-step explanation. Learn how to transform and differentiate y = [𝑡𝑎𝑛^(−1) ((3𝑥−𝑥^3)/(1−3𝑥^2 )] using calculus rules and trigonometric identities. Perfect for students and math enthusiasts looking to deepen their understanding of calculus concepts.
#Calculus #MathTutorial #Derivative #InverseTangent #Trigonometry #MathHelp #STEM #Education


welcome today we're going to learn about

logarithmic differentiation by finding

the derivative of yal e to power of E to

X this is a challenging function to

differentiate directly which is why

logarithmic differentiation is so useful

why do we use logarithmic

differentiation here well when we have

functions with exponents that contain

variables or other functions taking the

derivative directly can be very

complicated by using logarithms we can

transform our complex exponential into

something much easier to

differentiate let's solve the step by

step step one take the natural logarithm

of both sides of the equation this gives

us Ln of Y = Ln of e to power of E to X

step two

use the properties of logarithms to

simplify the right side remember when we

have Ln of something raised to a power

we can bring that power down in front of

the logarithm so this simplifies to Ln

of y = e to x * Ln of e and since Ln of

e equal 1 we get Ln of y = e to X

step three now we can differentiate both

sides with respect to X on the left side

we get 1/ y * dydx because we're using

the chain rule on the right side the

derivative of e to X is e to

X step four solve for

dydx multiply both sides by 1 Y which is

E power of E to X this gives us our

final answer Dy DX = E power of E to x +

x let's visualize what's happening with

the graph here's our original function y

= e power of E to X in blue and its

derivative in red notice how both fun

grow very rapidly as X increases with

the derivative growing even faster than

the original function in conclusion

logarithmic differentiation is a

powerful technique that helps us

differentiate complex exponential

functions by taking the natural log of

both sides using log properties and then

differentiating we can find derivatives

that would be very difficult to compute

directly remember this technique is AAL

useful when dealing with functions that

have variables and exponents or nested

exponentials thank you for watching

Mastering Logarithmic Differentiation: Finding d/dx[〖𝑒^𝑒〗^𝑥] Step by Step

Question: Find the derivative of y = [〖𝑒^𝑒〗^𝑥] using logarithmic differentiation.

📚 In this comprehensive calculus tutorial, we break down how to find the derivative of e(ex) using logarithmic differentiation. Perfect for AP Calculus, IB Mathematics, and college-level Calculus students!
🎯 What You’ll Learn:
• Why logarithmic differentiation is crucial for complex exponential functions
• Step-by-step solution process
• Properties of logarithms and their applications
• Chain rule implementation
• Common pitfalls and how to avoid them
📝 Follow along with our step-by-step solution!
#Calculus #Mathematics #LogarithmicDifferentiation #APCalculus #MathTutorial #CalcTutorial #IBMaths #MathEducation #LearnMath #ExponentialFunctions


let's solve this calculus problem step

by step we need to find the derivative

of tan x * sin inverse X using the

product rule the product rule states

that the derivative of a product of two

functions equals the first function time

the derivative of the second plus the

second function time the derivative of

the first let's write this out

mathematically in our case U = Tan x and

V = sin inverse X let's apply the

product rule to these functions now we

need to find the derivatives of sin

inverse X and tanx separately the

derivative of sin inverse X is 1 over

the < TK of 1 - x^2 and the derivative

of tan X is secant 2 x let's substitute

these derivatives back into our equation

we get tan x * 1 < TK 1 - x^2 plus s

inverse x * SEC 2 x and this gives us

our final answer the derivative of tan x

* sin inverse X = Tan x over < TK of 1 -

x^2 + sin inverse x * secant 2 x let's

summarize what we did we used the

product rule to split the derivative

into two terms then we found the

derivatives of sin inverse X and Tan x

separately and substituted them back

into our equation to get our final

answer

Mastering Calculus: Differentiating Products of Trigonometric & Inverse Functions[tan𝑥 .𝑠𝑖𝑛^(−1)𝑥]

Question: Differentiate the product [tan𝑥 .𝑠𝑖𝑛^(−1)𝑥]  with respect to  using the product rule.
Unlock the secrets of calculus with our step-by-step guide to differentiating products of trigonometric and inverse functions. Dive into the product rule, understand the process, and enhance your calculus skills. Perfect for students and math enthusiasts alike!
#Calculus #Differentiation #ProductRule #MathTutorial #Trigonometry #InverseFunctions


let's solve this calculus problem step

by step we need to find the derivative

of 5 e to x * log Bas of X the first

step is to rewrite log base of X using

the change of Base formula this states

that log base a of x equals log base e

of x / log base e of a which can also be

written as log Bas of e * log base e of

X for step 2 will apply the product rule

since we're multiplying e to X and the

natural log of x remember the product

rule states that the derivative of U * V

equal U Prime * V + U * V Prime for step

three we'll find the derivatives of each

component the derivative of the natural

log of x is 1 /x and the derivative of e

to X is just e to X step four let's

substitute these derivatives back into

our equation we get e to x * 1 /x plus

log base e of x * e to X for step five

we need to multiply everything by five

log base a of e which was our constant

from the change of Base formula this

gives us 5 e to x * 1 /x * log Bas uh of

e+ log Bas uh of e * log Bas e of X

finally in Step six we can simplify our

answer by recognizing that log base uh

of e * log base e of x equal log base of

X this gives us our final answer 5 e to

x * the Quan 1 /x * log base uh of e

plus log basa of x

and that's how we find the derivative of

5 e to the x * log base a ofx the key

steps were using the change of Base

formula applying the product rule and

then simplifying using logarithm

properties thank you for watching

Finding the Derivative of 5e^x log_a(x)  Using the Product Rule | Step-by-Step Explanation

In this video, discover how to differentiate 5e^x log_a(x) using the Product Rule in calculus.

What You'll Learn:
- Definition of the Product Rule
- Step-by-step differentiation of 5e^x log_a(x)
- Formula breakdown and example problem-solving

This example is ideal for calculus beginners or anyone reviewing the Product Rule for differentiation!
💡 Don't forget to LIKE, SHARE, and SUBSCRIBE for more math tutorials!
#Calculus #ProductRule #Differentiation #Derivative #MathTutorial #LearnMath #MathEducation #MathHelp


let's solve this step by step we need to

differentiate e to the power of 5 * the

natural logarithm of tangent of 5x

before we start differentiating let's

recall some important properties first

for exponents we know that e to the Ln

of x = x we'll also need the chain rule

which states that the derivative of a

composite function is the derivative of

the outer function times the derivative

of the inner function let's start by

simplifying our expression using the

property of exponents notice that we

have e to the power of 5 Ln of tangent

5x using our property that e to the Ln

of x = x this simplifies to tangent of

5x to 5th power now we can differentiate

this expression the outer function is

the power function and the in function

is tangent of 5x using the chain rule

we'll multiply the derivative of the

outer function by the derivative of the

inner function let's focus on finding

the derivative of tangent of 5x remember

that the derivative of tangent is secant

squar but since we have 5x inside we

need to multiply by the derivative of 5x

which is just 5 now let's put everything

together

we multiply 5 * tangent of 5x 4 power *

secant 5x this simplifies to 25 tangent

of 5x 4th power * secant SAR of 5x if

you prefer we can write this in an

alternative form using the fact that

secant s = 1 + tangent 2 this would give

us 25 tangent of 5x 4 power * the 1 +

tangent 5x let's review what we did we

first simplified the expression using

properties of exponents and logarithms

then we applied the chain rule twice

once for the power function and once for

the tangent function finally we combine

terms to get our answer

Step-by-Step Guide to Differentiating Exponential Logarithmic Functions 𝑒^(5 ln(𝑡𝑎𝑛 5𝑥))

In this tutorial, we provide a detailed step-by-step guide to differentiating the expression 𝑒^(5 ln(𝑡𝑎𝑛 5𝑥)) . Learn about the fundamental properties of exponents and logarithms, and how to apply the chain rule effectively. Whether you’re a calculus student or just brushing up on your math skills, this video will walk you through each calculation clearly and concisely!
#Calculus #Differentiation #MathTutorial #ExponentialFunctions #LogarithmicFunctions #ChainRule #MathEducation


welcome to this calculus tutorial on

finding derivatives of logarithmic

functions today we'll solve a complex

problem involving logarithms with

different bases let's break this down

step by step first let's recall the

change of Base formula for any logarithm

log base uh of xals the natural log of x

/ the natural log of a using this

formula let's rewrite our original

expression the first term log base of X

becomes natural log of x / natural log

of a for the second term log base X of a

it becomes natural log of a divided

natural log of x now we can

differentiate each term separately using

the quotient rule let's start with the

first term remember the quotient rule

states that the derivative of U over V

equal V * the of uus U * the derivative

of V all over V ^2 for the first term U

is natural log of x and V is natural log

of a the derivative of natural log of x

is 1 /x and the derivative of natural

log of uh is zero since is a constant

for the second term U is natural log of

a and V is natural log of x again the

derivative of natural log of a is zero

and the derivative of natural log of x

is 1 /x now we can combine these terms

using the sum rule the derivative of our

original expression is 1 /x Ln of a

minus Ln of a overx * Ln of x^2 and

that's our final answer let's review

what we did we used the change of Base

formula to convert to Natural logarithms

applied the quotient rule to each term

and combined our results using the sum

rule this problem demonstrates the

importance of knowing your logarithm

properties and differentiation rules

thank you for watching

Complex Logarithmic Differentiation: Solving dy/dx for loga(x) + logx(a) | Calculus Tutorial

Question:
Find the derivative of y = loga(x) + logx(a)

📚 Master complex logarithmic differentiation in this comprehensive calculus tutorial! Learn how to find the derivative of y = loga(x) + logx(a) step by step.
In this video, we cover:
⭐ Breaking down complex logarithmic expressions
⭐ Change of base formula applications
⭐ Chain rule and quotient rule implementation
⭐ Natural logarithm conversions
⭐ Sum rule for derivatives
⭐ Final simplification techniques
🎯 Perfect for:
• High school AP Calculus students
• College Calculus I & II students
• Math majors
• Students preparing for exams
#Calculus #Mathematics #LogarithmicDifferentiation #MathTutorial #CalcTutorial #MathHelp #APCalculus #CollegeMath #Derivatives #MathEducation


let's solve this challenging derivative

problem step by step we need to find the

derivative of log base cine X of tangent

X the first step is to convert this to

Natural logarithms using the change of

Base formula the change of Base formula

states that log base uh of b equals

natural log of B divided by natural log

of a applying this to our problem we get

natural log of tangent x / natural log

of cine X now we need to take the

derivative of this quotient we'll use

the quotient rule which states that the

derivative of U over V equal V * the

Dera of uus U * the derivative of V all

over V ^2 let's apply the quotient rule

to our expression here U is natural log

of tangent X and V is natural log of

cine X this gives us natural log of

cosine x times the derivative of natural

log of tangent x minus natural log of

tangent x * the derivative of natural

log of cine X all over natural log of

cine x^ 2 now we need to find the

derivatives of the natural logarithms

the derivative of natural log of tangent

x = 1 / tangent x * the derivative of

tangent X which is secant 2 x and the

derivative of natural log of cine x = 1

/ cosine x * sinx let's substitute these

derivatives back into our quotient rule

expression after substituting and

simplifying we get natural log of cine x

* cosine X over sin x * cosine 2 X Plus

natural log of tangent x * sin x cosine

X all over natural log of cosine

x^2 finally we can express this using

secant and cosecant our final answer is

secant x * cose x * natural log of cine

X Plus natural log of tangent x *

tangent X all over natural log of cine

x^2 this problem demonstrated the use of

several calculus Concepts the change of

Base formula for logarithms the quotient

rule the chain rule

and trigonometric identities thank you

for watching

Derivatives of Logarithms with Trigonometric Bases [log_𝑐𝑜𝑠𝑥(tan𝑥)] | Advanced Calculus

Question:
Find the derivative of [log_𝑐𝑜𝑠𝑥(tan𝑥)] with respect to x.

Master the art of differentiating complex logarithmic expressions with trigonometric bases! In this comprehensive tutorial, we solve step-by-step:
• How to differentiate log_cos(x)(tan x)
• Using the change of base formula
• Applying the quotient rule
• Working with trig derivatives
• Simplifying complex trig expressions
Perfect for:
📚 AP Calculus BC students
🎓 College Calculus II students
📖 Math majors
✏ Test preparation
#Calculus #Mathematics #Derivatives #MathTutorial #CalculusBC #CollegeMath #LogarithmDerivatives #TrigonometryCalculus


in this video we'll solve a challenging

differentiation problem involving

inverse trigonometric functions let's

break it down step by step here's our

problem find the derivative of yal

inverse tangent of 6 < tkx over 1 - 9x

this looks complicated at first glance

but we can solve it using substitution

and trigonometric identities let's start

with a clever sub subtitution we'll let

3 < tkx equal tangent T this means tal

inverse tangent of 3 rootx now let's

rewrite our expression using the

substitution the numerator 6 rootx can

be written as 2 * 3 < tkx which is 2 Tan

t and in the denominator 1 - 9x is

equivalent to 1 - tangent 2 T because 9x

is s of 3 3un X which we set equal to

tan T here's where we use a key

trigonometric identity the expression 2

tan t/ 1 - tan^ 2 T is equal to tan of 2

T therefore y equals inverse tangent of

tan of 2T which simplifies to Just 2 T

now we can substitute back our

expression for t y = 2 * inverse tangent

of of 3 < tkx to find dydx we'll use the

chain rule remember the derivative of

inverse tangent is 1/ 1 + U ^2 where U

is the inside function then we multiply

by the derivative of the inside function

let's simplify this step by step first 1

+ 3un x^2 = 1 + 9x the derivative of 3 <

tkx is 3 over 2un X finally when we

multiply everything out and simplify we

get our answer 3/ < tkx * 1 + 9x and

that's our solution we use substitution

trigonometric identities and the chain

rule to find this derivative thanks for

watching

Mastering Derivatives: Solving [𝑡𝑎𝑛^(−1)  (6√𝑥  )/(1−9𝑥)] Step-by-Step

Question:
Find the derivative of  y=[𝑡𝑎𝑛^(−1)  (6√𝑥  )/(1−9𝑥)] in terms of x .
Dive into calculus with this step-by-step explanation of how to find the derivative of y=[𝑡𝑎𝑛^(−1)  (6√𝑥  )/(1−9𝑥)]. Learn how to apply trigonometric identities and calculus rules like the chain rule to solve this problem easily. Perfect for students and math enthusiasts!
#Calculus #Derivatives #Trigonometry #MathTutorial #ChainRule #InverseTrigonometricFunctions


let's solve this differentiation problem

step by step we'll start with a question

first we need to understand what X De

means in terms of radians when we see X

De we can convert it to radians by

multiplying X by pi over

180 so our expression becomes the

derivative of x piun over 180 * cine of

PX over

180 now this is a product of two

functions so we'll need to use the

product rule the product rule states

that the derivative of U * v = u * the

derivative of V plus v * the derivative

of U applying the product rule we get x

pi over 180 * the derivative of cosine

pix over

180 plus cosine pix over 180 * the

derivative of x piun over

180 now let's handle the derivative of

cosine using the chain rule the

derivative of cosine is negative s and

we multiply by the derivative of the

inner function the derivative of x piun

over 180 is simply pi over

180 finally we can factor out pi over

180 to get our final answer pi over 180

* the quantity cosine pix over 100 and

80 - pkx over 18 80 * sin pix over

180 and that's our complete solution we

use both the product rule and chain rule

along with our knowledge of

trigonometric derivatives to solve this

problem

Understanding Calculus: Differentiating Trigonometric Functions (𝑥° 𝑐𝑜𝑠𝑥° ) Step by Step

Learn how to differentiate complex trigonometric expressions with ease. In this video, we’ll walk you through a step-by-step breakdown of differentiating the function (𝑥° 𝑐𝑜𝑠𝑥° )  , highlighting key calculus concepts such as the product rule, chain rule, and unit conversions from degrees to radians. Perfect for students looking to deepen their understanding of derivatives and improve their calculus skills!
#Calculus #MathTutorial #Differentiation #Trigonometry #LearnMath


welcome today we're going to learn about

logarithmic differentiation by finding

the derivative of y = x to X power this

is a challenging function to

differentiate directly but logarithmic

differentiation makes it much more

manageable let's see what makes this

function tricky here's what y = x to the

X looks like notice how it's different

from regular power functions because the

exponent changes as X changes step one

let's start by taking the natural

logarithm of both sides this gives us

natural log of Y = natural log of x x

step two using the logarithm property

that the log of a power equals the

exponent time the log we can simplify

this to natural log of y = x * natural

log of x step three now we can

differentiate both sides with respect to

X on the left side we get 1/ y * dydx

because we're using the chain rule on

the right side we need to use the

product rule step four let's apply the

product rule to the right side the

derivative of x * natural log of x = x *

1x + natural log of x * 1 this

simplifies to 1 + natural log of x

step five now we can solve for

dydx multiply both sides by Y and we get

Dy DX = y * the 1 + natural log of x

step six finally we can substitute back

y = x to X this gives us our final

answer the derivative of x to x = x to

the x * thean 1 + natural log of x let's

visualize what this derivative looks

like here's the original function in

blue and its derivative in red notice

how the derivative grows much faster

than the original function logarithmic

differentiation is particularly useful

when dealing with functions that have

variables in the exponent or when

dealing with products of multiple terms

it allows us to convert these complex

Expressions into simpler ones using the

properties of logarithms before

differentiating and that's how we use

logarithmic differentiation to find the

derivative of x to the X power thanks

for watching

Master Logarithmic Differentiation: Solving [𝒚=𝒙^𝒙 ] Step by Step”

Question: Find the derivative of y = x^x using logarithmic differentiation.

📚 In this comprehensive tutorial, we break down how to find the derivative of y = x^x  using logarithmic differentiation. Perfect for calculus students and math enthusiasts!
What you’ll learn:
✅ Why logarithmic differentiation is essential for complex exponential functions
✅ Step-by-step solution process
✅ Key logarithm properties and their applications
✅ Product rule implementation
✅ Common pitfalls and how to avoid them
⭐ Don’t forget to like and subscribe for more calculus tutorials! ⭐
#Calculus #Mathematics #LogarithmicDifferentiation #MathTutorial #CalcTutorial #MathEducation #LearnMath #Derivatives


welcome we'll find the derivative of e

to the power of MX using first

principles this is a fundamental

derivative that's important to

understand let's start with our function

let F ofx equal e to the power of MX

step one is to find f of x + H when we

substitute x + H into our function we

get f of x + h = e to the power of M *

thean x +

H step two we'll use the definition of

the derivative the derivative is defined

as the limit as H approaches zero of f

of x + H - FX all divid H let's

substitute our

Expressions step three let's simplify

this expression first we can rewrite e

to the power of M * x + H as e to the

power of MX * e to the power of MH then

we can factor out e to the power of MX

from the numerator step four now comes

the key Insight we'll use the series

expansion for e to the power of MH E

power of MH = 1 + m

H 1 factorial + m^ 2 h^ 2/ 2 factorial +

m Cub H cubed over 3 factorial and so

on step five let's substitute this

series into our limit when we subtract

one and divide by H we get e to the

power of MX * the limit as H approaches

Zer of m + m^ 2 H over 2 Factor + MB h^2

3 factorial and so

on step six finally as H approaches zero

all terms with h go to zero this leaves

us with just m * e to the power of MX

therefore the derivative of e to the

power of MX = M * E power of MX this is

a powerful result that shows us that the

exponential function is special it's the

only function that is proportional to

its own derivative when m = 1 we get the

even more special case where the

function exactly equals its own

derivative let's visualize this function

and its derivative the blue curve shows

e to the power of MX and the red curve

shows its derivative M * E power of MX

notice how they're proportional to each

other mother

Understanding the Derivative of e^mx  Using First Principles

Dive into the fundamentals of calculus as we explore how to determine the derivative of the exponential function e^mx  using first principles. Follow along with step-by-step explanations and examples to enhance your understanding of derivatives and exponential functions.
#Calculus #Derivatives #ExponentialFunctions #MathTutorial #FirstPrinciples #Defferentiation


let's learn how to find the derivative

of a natural logarithm containing a

quadratic expression let's start with

our expression we want to find the

derivative of Ln of ax^2 + BX + C with

respect to X to solve this we'll use the

chain rule for logarithmic

differentiation the derivative of

natural log of U with respect to X = 1/

U * the derivative of U with respect to

X let's apply this to our problem in our

case U = ax^2 + BX + C so we can write

our derivative as 1 over the quadratic

expression times the derivative of that

same expression now let's focus on

finding the derivative of the quadratic

expression using the power rule and the

sum rule the derivative of ax^2 is 2 ax

the derivative of BX is B and the

derivative of the constant C is zero

combining these terms the derivative of

our quadratic expression is 2 ax + B

finally we can write this as a single

fraction our final answer is 2 ax + B

over ax^2 + BX + C let's visualize with

a specific example let's say a = 1 b = 2

and c = 1 this means we're finding the

derivative of the natural log of x^2 +

2x + 1 and its derivative would be 2x +

2/ x^2 + 2x + 1 notice how the

derivative gives us the rate of change

of the logarithmic function at each

point in summary we use the chain rule

for logarithmic differentiation to find

the derivative of a natural logarithm

containing a quadratic expression the

key steps were identifying our U

function applying the chain Rule and

then using the power rule to

differentiate the quadratic expression

Logarithmic Differentiation (𝑙𝑛(𝑎𝑥^2+𝑏𝑥+𝑐)) Step by Step Guide

Unlock the secrets of calculus with our in-depth tutorial on differentiating logarithmic functions. In this video, we’ll guide you through the process of finding the derivative of  (𝑙𝑛(𝑎𝑥^2+𝑏𝑥+𝑐))   using the chain rule. Perfect for students and math enthusiasts looking to enhance their understanding of calculus concepts.
#Calculus #Logarithms #Differentiation #MathTutorial #ChainRule


let's solve this calculus problem step

by step we need to differentiate the

natural logarithm of the sum of two

square roots let's break this down into

manageable steps using the chain rule

first recall that the derivative of

natural log is 1 over the function times

the derivative of the inside function

applying this to our problem we get 1

over the sum of our square roots times

the derivative of that sum

next we can split up the derivative of

the sum into the sum of the derivatives

this is possible because the derivative

is a linear operator now let's recall

the power rule for square roots the

derivative of square root of U is 1/ 2 *

the square < TK of U * the derivative of

U applying this to both terms we get

this more complex expression notice how

we're applying the chain rule again to

each square root the derivatives of x -

2 and x + 1 are both simply one which

simplifies our expression to this now

comes the clever part we can find a

common denominator inside the curly

brackets when we do this the numerator

becomes the sum of square roots which

matches our denominator outside the

brackets and finally when we multiply

these fractions the sum of of square

roots cancels out in the numerator and

denominator leaving us with our Final

Answer 1 over 2 * the product of our

square roots this problem showcases

several important calculus Concepts the

chain rule the derivative of natural

logarithm the derivative of square roots

and clever algebraic manipulation to

simplify our result

How to Differentiate Logarithmic Functions [𝒍𝒏(√(𝒙−𝟐)+√(𝒙+𝟏))] with Square Roots: Step-by-Step Guide

Question:
Differentiate the expression  [𝒍𝒏(√(𝒙−𝟐)+√(𝒙+𝟏))] with respect to x .

Are you struggling with differentiating logarithmic functions involving square roots? In this video, we break down the process of differentiating the expression [𝒍𝒏(√(𝒙−𝟐)+√(𝒙+𝟏))]  step-by-step. Follow along as we use the chain rule and simplify the derivative to achieve clarity and accuracy. Perfect for students preparing for calculus exams or anyone needing a refresher. Don’t forget to subscribe for more math tutorials!

#Calculus #Differentiation #Logarithms #SquareRoots #MathTutorial #ChainRule


let's solve this complex trigonometric

derivative step by step we'll start by

understanding what we're trying to

differentiate our problem is to find the

derivative of secant inverse of the

quantity 1 + x^2 over 1 - x^2 with

respect to X to solve this we'll use a

clever substitution let's let xal

tangent of t this means tal inverse

tangent of x our first step is to use

the relationship between secant inverse

and cosine inverse remember that secant

inverse of Y = cosine inverse of 1 / y

now let's substitute X = tangent T into

our expression this gives us cosine

inverse of 1 - tangent 2 T over 1 +

tangent 2 T here's where we need to

recall a key trigonometric identity the

expression 1 - Tang 2 T over 1 + tangent

2 T is equal to cosine of 2T this

simplifies our expression to cosine

inverse of cosine 2T which equals 2T now

we need to differentiate 2 t with

respect to X remember that t equals

inverse tangent of X so we'll use the

chain rule the derivative of inverse

tangent of X is 1/ 1 + x^2 therefore our

final answer is 2 over 1 + x^2 let's

verify this makes sense our answer is

always defined except when x = + or

minus1 which is exactly where our

original expression would be undefined

due to division by 0o in the denominator

1us x^2

Derivative of Inverse Secant: Step-by-Step Explanation [𝑠𝑒𝑐^(−1)   (1+𝑥^2)/(1−𝑥^2 )]

Unlock the secrets of calculus with our detailed step-by-step guide on differentiating the inverse secant function! This video breaks down the process of finding the derivative of [𝑠𝑒𝑐^(−1)   (1+𝑥^2)/(1−𝑥^2 )] using trigonometric identities and substitution methods. Follow along for a clear and insightful explanation that will enhance your understanding of calculus concepts.
#Calculus #Derivative #InverseSecant #Trigonometry #MathTutorial


let's solve this calculus problem step

by step we need to find the derivative

of x to the 5th power time log base of

of X the first step is to rewrite log

base uh of X using the change of Base

formula this states that log Bas uh of x

equals log base uh of e * log base e of

X next we can factor out log base uh of

e since it's a constant this gives log

base uh of e * the derivative of x to

5th power * natural log of x now we can

apply the product rule the product rule

states that the derivative of U * v = u

* the derivative of V plus b * the

derivative of U let's evaluate each

derivative separately the derivative of

natural log of x is 1 /x and the

derivative of x to the 5th power is 5X

4th power substituting these derivatives

back into our equation we get log base

of e * X to 5 * 1 /x plus natural log of

x * 5x 4th now let's simplify x 5 * 1 /x

simplifies to X 4th and then we have +

5x 4 * natural log of X for our final

step we can factor out x to the 4th

power this gives us X to 4th times the

quantity log base uh of e + 5 log base

of X and there we have it we found the

derivative of x to the 5th power * log

base a of X by using the change of Base

formula the product rule and some

careful algebraic manipulation

Finding the Derivative of 𝑥^5.log_a(x)  Using the Product Rule | Step-by-Step Explanation.

In this video, learn how to find the derivative of 𝑥^5.log_a(x)
 using the Product Rule in calculus. 🎓
📌 What You'll Learn in This Video:
What is the Product Rule?
Step-by-step solution to differentiate 𝑥^5.log_a(x)
Formula breakdown and example problem-solving.
This example is perfect for beginners in calculus or anyone reviewing the Product Rule for differentiation!
💡 Don't forget to LIKE, SHARE, and SUBSCRIBE for more math tutorials!
#Calculus #ProductRule #Differentiation #Derivative #MathTutorial #LearnMath #MathEducation #MathHelp


let's solve this differentiation problem

step by step we start with the

expression square root of 1 over e to X

step one let's rewrite this expression

in a form that's easier to differentiate

we can rewrite this as e tox to the^ of

12 this is because the square root is

the same as raising to the power of 12

and 1/ E to X can be written as e to X

step two now we'll use the chain rule to

differentiate this expression the chain

rule states that when differentiating U

to the N we get n * U the

nus1 time the derivative of U step three

let's identify our U and N here U is e

to thex and N is 12 step four let's

apply the chain rule we multiply 12 * e

thex to the power of -2 * the derivative

of e to thex which is e to thex step

five now let's simplify this expression

first we can rewrite e to thex to the

power of -2 as 1 over the < TK of e

tox then we can combine the terms with e

to

thex finally we combine the exponents

and simplify to get our final answer

-1/2 * the < TK of e to X and that's how

we differentiate the square TK of 1/ e

to X we use the chain Rule and careful

algebraic manipulation to arrive at our

answer

9.3-2.2 Mastering the Chain Rule: Differentiating (√(1/𝑒^𝑥)  Step-by-Step

Unlock the secrets of differentiation with our in-depth tutorial on finding the derivative of (√(1/𝑒^𝑥) . Follow along as we guide you through each step, using the chain rule to achieve a clear understanding of this complex mathematical process. Perfect for students and math enthusiasts looking to enhance their calculus skills!
Hashtags:
#cpmrana #MathTutorial #Calculus #ChainRule #Differentiation #ExponentialFunctions #hsc #hschighermath #1stpaper #hschighermath1stpaper #englishversion


let's work through this complex

derivative step by step we're going to

find the derivative of the square OT of

e to the power of square < TK X let's

break this down into its component parts

to better understand what we're dealing

with we have three nested functions here

an outer square root then an exponential

function e and finally another square

root of x to solve this we'll need to

use the chain rule multiple times the

chain rule tells us that when

differentiating composite functions we

multiply the derivative of the outer

function by the derivative of the inner

function let's start with the outer

function the derivative of square root

of U with respect to U is 1/ 2 * the

square < TK of U in our case U is e^ of

square < TK of X next we need to find

the derivative of the middle function

the derivative of e to the power of U

with respect to U is e to the power of U

but remember r u here is square < TK of

X finally let's find the derivative of

the innermost function the derivative of

square < TK X is 1 2 * the < TK X now

let's put it all together using the

chain rule we multiply all three parts 1

over 2 * theare < TK of e to power of <

TK x * E power of < TK x * 1/ 2 * the <

TK X finally let's simplify this

expression the numerator Becomes e^ of <

TK of X and the denominator becomes 4 *

the < TK of x * the < TK of e to power

of < TK X and that's our final answer we

use the chain rule multiple times to

break down this complex derivative into

manageable steps and then combined

everything back together

9.3-2.3 Mastering Calculus Differentiating Composite Functions (√𝑒^√𝑥)  with Chain Rule

In this video, we break down the process of differentiating complex composite functions using the chain rule. Learn how to tackle expressions involving square roots and exponential functions step by step. Perfect for students and anyone looking to strengthen their calculus skills!
#hsc #hschighermath #1stpaper #hschighermath1stpaper
#englishversion #Calculus #ChainRule #Derivatives #MathTutorial #CompositeFunctions


welcome today we'll learn how to find

the derivative of a complex function

using logarithmic differentiation our

question is to find the derivative of y

= sin x raed to X power let's solve the

step by step first let's say y = sin x

raed to X power this is our original

function to use logarithmic

differentiation we take the natural log

of both sides using the logarithm

property that log of a power equals the

power time the log of the base we can

simplify this to x * Ln of sinx now we

can differentiate both sides with

respect to X on the right side we need

to use the product rule since we have a

product of two terms applying the

product rule we get x * the derivative

of Ln sinx X Plus Ln sign x * the

derivative of x on the left side we use

the chain rule the derivative of Ln Y is

1/ y * Dy DX on the right side we use

the chain rule again for the derivative

of Ln sinx which gives us cin X over

sinx now we can multiply both sides by y

to isolate

dydx and we can substitute back y = sin

x to X power finally we can recognize

that cine x/ sinx is the coent of X and

there we have it we've found the

derivative using logarithmic

differentiation this method is

particularly useful when dealing with

functions that have variable exponents

the key steps were one taking the

natural log of both sides two using

logarithm properties 3 using the product

rule 4 applying the chain Rule 5 and

finally substituting back our original

function

Mastering Logarithmic Differentiation: Solving [(sin x)^x] Step by Step

QUESTION:
Find the derivative of y=[(sin x)^x]   using logarithmic differentiation.

Learn how to find the derivative of (sin x)^x using logarithmic differentiation in this comprehensive calculus tutorial. We break down this challenging problem into manageable steps, showing you:
• How to apply natural logarithms strategically
• Using logarithm properties effectively
• Implementing the product rule and chain rule
• Working with trigonometric functions
• Simplifying using trigonometric identities
Perfect for calculus students and anyone preparing for AP Calculus, IB Mathematics, or college-level calculus courses.
#Calculus #Mathematics #LogarithmicDifferentiation #ChainRule #ProductRule #MathTutorial #CalulusHelp #APCalculus #IBMaths #MathEducation


welcome today we'll solve a challenging

derivative problem involving inverse

tangent let's find the derivative of

this complex expression let's break this

down step by step first let's let y

equal our expression and try to simplify

what's inside the inverse tangent our

first step is to factor out from both

the numerator and denominator we can

rewrite the fraction as a Time 1 + BX

over a / a Time 1 - bx/ A the A's cancel

out in the fraction giving us this

simpler form now here's where we use a

key Insight this form can be broken down

using the inverse tangent addition

formula when we have a fraction in this

form it equals the sum of inverse

tangent of 1+ inverse tangent of the

remaining term we know that inverse

tangent of 1 = pi

4 so we can simplify this further now we

can differentiate the derivative of a

constant pi/ 4 is zero and for the

inverse tangent term we'll use the chain

rule let's evaluate the derivative of BX

over a which is simply B over a now

we'll work on simplifying this fraction

first let's get a common denominator in

the first fraction when we have a

complex fraction like this this we can

flip the denominator and

multiply finally we multiply these

fractions together and get our final

answer therefore the derivative of

inverse tangent of a plus BX over a

minus BX is equal to a / a 2 + B ^2 x^2

this was a challenging problem that

required us to use the inverse tangent

addition formula and careful algebraic

manipulation the the key was breaking it

down into manageable steps

Calculus Explained: Derivative of Inverse Tangent Function [𝑡𝑎𝑛^(−1)((𝑎+𝑏𝑥)/(𝑎−𝑏𝑥))]

Question:
Find the derivative of y=[𝑡𝑎𝑛^(−1)((𝑎+𝑏𝑥)/(𝑎−𝑏𝑥))].

Dive into calculus with this step-by-step guide on differentiating the function y=[𝑡𝑎𝑛^(−1)((𝑎+𝑏𝑥)/(𝑎−𝑏𝑥))]. Learn how to use trigonometric identities and the chain rule to find the derivative efficiently. Perfect for students and math enthusiasts looking to enhance their understanding of advanced calculus concepts.
#Calculus #Derivative #Trigonometry #MathTutorial #InverseTangent #ChainRule


let's solve this calculus problem step

by step we need to find the derivative

of f ofx = x to the n * natural log of

2x to solve this we'll use the product

rule the product rule states that the

derivative of U * v = u * the derivative

of V + V * the derivative of U in our

case u = x to the n and V equal natural

laog 2x let's apply the product rule we

start with x n * the derivative of

natural log of 2x plus natural log of 2x

* the derivative of x to the N now let's

find each derivative separately the

derivative of natural log of 2x is 1 /

2x * 2 which simplifies to 1 /x and the

derivative of x to the N is n * x n -1

substituting these back in we get x n *

1 2x * 2 plus natural log of 2x * NX n

-1 let's simplify the first term x n * 2

/ 2x simplifies to X N /x which is X N

-1 finally we can factor out x to the

nus1 from both terms this gives us our

final answer x to the nus 1 * the

quantity 1 + n * natural log of 2x to

verify this is correct we can check that

our answer contains all the terms we

expect from the product rule and that we

follow the chain rule correctly when

differentiating natural log of 2x this

completes our solution

Mastering Calculus: Differentiating (𝑥^𝑛  ln2𝑥 )  Step-by-Step

Unlock the secrets of calculus with our detailed step-by-step guide on differentiating (𝑥^𝑛  ln2𝑥 )  . In this video, we’ll walk you through the process using the product and chain rules, ensuring you understand each step. Perfect for students and calculus enthusiasts aiming to deepen their understanding. Subscribe for more math insights!
#Calculus
#MathTutorial
#Differentiation
#LearnMath
#StudyHelp


welcome today we'll solve a challenging

differentiation problem using

logarithmic

differentiation our task is to find the

derivative of y = s < TK X raed to the^

of square otk of X when we see an

expression where the variable appears in

both the base and exponent logarithmic

differentiation is often the best

approach let's start by letting y equal

our expression

and then take the natural logarithm of

both sides using the logarithm property

that states the log of a power equals

the exponent time the log of the base we

can rewrite this as Square < TK of x *

the natural log of square < TK of X now

we can simplify the natural log of

square < TK of X remember the natural

log of x to any power equals that power

time the natural log of x so Ln of s T

of x = 12 Ln of X here's where the magic

of logarithmic differentiation comes in

we can now differentiate both sides with

respect to X on the left side we get 1

over y *

dydx on the right side we need to use

the product rule since we're

differentiating the product of square

root of x and natural log of x this

gives a square root of x * the

derivative of Ln X Plus Ln x * the

derivative of < TK of X now let's

evaluate these derivatives the

derivative of lnx is 1 /x and the

derivative of sare < TK X is 1 / 2 < TK

X simplifying the right side we get 1/ 2

* the sum of 1 / < TK of x + Ln x/ 2 <

TK X

to solve for dydx we multiply both sides

by y remembering that y = < TK X to^ of

< TK

X after some algebra we can combine like

terms and simplify this to get our final

answer 1/4 * < TK X to the^ of < TK of x

-1 * 2 + Ln X

and there we have it we've successfully

found the derivative using logarithmic

differentiation this technique is

particularly useful when dealing with

Expressions involving variables in both

the base and exponent

Mastering Logarithmic Differentiation: Solving (√x)^(√x) Step by Step | Advanced Calculus

Question: Find the derivative of y = (√x)^(√x) using logarithmic differentiation.

Learn how to find the derivative of (√x)^(√x) using logarithmic differentiation in this comprehensive tutorial. Perfect for calculus students and math enthusiasts!
In this video, we break down:
• Why logarithmic differentiation is necessary
• Step-by-step solution process
• Key logarithm properties
• Product rule application
• Final simplification techniques
📚 Prerequisites:
Basic differentiation rules
Natural logarithms
Chain rule
Product rule
#calculus #mathematics #logarithms #derivatives #mathtutorial #calculus101 #mathhelp #education


let's solve this differentiation problem

step by step we need to find the

derivative of x * the square root of

sinx with respect to X since this is a

product of two functions we'll use the

product rule remember the product rule

states that the derivative of U * v = u

Prime * V + U * V Prime in our case u =

x and V = the sare < TK of sinx let's

apply the product rule to our expression

first we'll find U Prime which is the

derivative of x that's simply one then

we multiply that by V which is square <

TK of sin x so our first term is square

otk of sin x for the second term we need

you which is x * V Prime which is the

derivative of square < TK of sinx to

find this we'll need to use the chain

rule the derivative of square < TK of U

is 1 / 2 < TK of U * U Prime so our

second term will be x * 1 / 2 < TK sin x

* the derivative of sin x and we know

the derivative of sinx is cine X now we

can combine our terms we have < TK sin x

+ x * 1 / 2 < TK sin x * Co cosin X and

finally we can write this as sare < TK

sin x + x cosine x / 2 < TK of sinx this

is our final answer let's review what we

did we used the product rule to split

the derivative into two terms then use

the chain rule to find the derivative of

square otk of sinx and finally combine

our terms to get our answer

Mastering Derivatives (𝑥√(sin𝑥 )): Step-by-Step Product and Chain Rule Explained

Unlock the secrets of calculus with our comprehensive video on differentiating the expression (𝑥√(sin𝑥 )) . Follow along as we break down each step using the product and chain rules. Perfect for students and math enthusiasts looking to sharpen their skills! 📚✏
#Calculus #MathTutorial #Differentiation #LearnMath #ProductRule #ChainRule


let's solve this calculus problem step

by step the question asks us to

differentiate the function f of xal 3

raised to the inverse s of X let's break

this down using a step-by-step approach

we'll use a two column format to show

both the mathematical steps and their

explanations first let's identify what

we're dealing with this is a composite

function where we have three Rays to a

power and that power is itself the

inverse sign of X we'll need to use the

chain rule here the chain rule tells us

that when differentiating an exponential

function a to the U we get a to the U *

the natural log of times the derivative

of U applying this to our problem we get

3 to the inverse sin of X time the

natural log of 3 times the derivative of

inverse sin of X now we need to recall

that the Der ative of inverse sin of X

is 1 over the < TK of 1 - x^2

substituting this in we get our final

answer 3 to the inverse sign of x *

natural log of 3 all over the square <

TK of 1 - x^2 let's verify our answer

makes sense the domain of this

derivative will be the same as the

domain of inverse sign which is -1 to to

positive 1 also notice that as X

approaches either 1 or - 1 our

denominator approaches zero which means

the derivative grows very large at these

points just as we would expect and that

completes our solution remember the key

steps identify the composite function

apply the chain Rule and recall the

derivative of inverse sign

Differentiating Exponential Functions Solving [3^(𝑠𝑖𝑛^−1 𝑥)]

Learn how to differentiate the function [3^(𝑠𝑖𝑛^−1 𝑥)] step-by-step in this comprehensive tutorial. We’ll guide you through applying the chain rule and simplifying complex expressions involving inverse trigonometric functions. Mastering these calculus skills is essential for any math enthusiast or student.
#Calculus #Differentiation #MathTutorial #ExponentialFunctions #ChainRule


let's solve this challenging derivative

problem step by step we'll find the

derivative of inverse tangent of 4X over

1 -

4x^2 this is a complex expression that

would be difficult to differentiate

directly let's try a substitution

approach let's say that 2x = tangent of

t therefore t equal inverse tangent of

now let's look at our original

expression we can rewrite 4X over 1 -

4x^2 as 2 *

2x/ 1 -

2x^2 now we can substitute tangent t for

2x this gives us 2 * tangent t/ 1 - Tang

2 T this is a key Insight this

expression is actually equal to tangent

of 2T this is the double angle formula

for tangent so our original expression

becomes inverse tangent of tangent of 2T

which simplifies to Just 2 T now

remember that t equals inverse tangent

of 2x so we can substitute back to get 2

* inverse tangent of 2x now we can

differentiate this expression using the

chain rule we get 2 * the derivative

of inverse tangent of 2x the derivative

of inverse tangent is 1/ 1 + x^2 and we

need to multiply by the derivative of

the inside function this simplifies to 4

over 1 + 4x^2 which is our final answer

this problem demonstrates the power of

substitution and trigonometric

identities in calculus by making a

clever substitution we transform a

complex derivative into a much simpler

one

How to Calculate the Derivative of [𝑡𝑎𝑛^(−1)   (4𝑥 )/(1−4𝑥^2 )] - Step-by-Step Explanation!

Question
Find the derivative of [𝑡𝑎𝑛^(−1)   (4𝑥 )/(1−4𝑥^2 )] .
Unlock the secrets to finding the derivative of  [𝑡𝑎𝑛^(−1)   (4𝑥 )/(1−4𝑥^2 )]  with this in-depth tutorial! We’ll guide you through each math step and provide clear explanations, ensuring you grasp the concept fully. Perfect for high school and college students tackling calculus! Don’t forget to like, subscribe, and comment with any questions.
#Calculus #Derivatives #InverseTangent #MathTutorial #StepByStepMath


let's solve this calculus problem step

by step using the quotient rule we need

to find the derivative of the function f

of x = tangent x + Cent X all divided 3

e to X first let's recall the quotient

rule the derivative of U over V equal V

* the derivative of U minus U * the

derivative of V all over V ^2 let's

identify our u and v fun funs here U =

tangent x + Cent X and V = 3 e to X now

let's find their derivatives the

derivative of tangent X is secant 2 x

and the derivative of Cent X is Nega

cose 2 x so the derivative of U is

secant 2 xus cose 2 x the derivative of

V is simply 3 e to the X now let's

substitute everything into the quotient

rule formula we get 3 e to the x *

secant 2 x - cose 2 x - tangent x + Cent

x * 3 e to X all over 3 e x^2 let's

simplify the numerator first we can

Factor Out 3 e to X this gives us 3 e to

x * the quantity C 2 x - cose 2 x -

tangent x - x all over 3 e to x^2 next

recall that secant 2 x = 1 + Tang 2 x

and cose 2 x = 1 + cang 2 x substituting

these in we get tangent s x - Cent 2 x -

tangent x - cang X finally we can Factor

this expression notice that tangent x +

Cent X is a common factor when we Factor

it out we get tangent x + Cent x *

tangent x - Cent x -1 all over 3 e to X

to verify this is correct we can check

that this matches our original problem's

derivative the quotient rule has helped

us transform a complex fraction into a

more manageable form

Differentiating Trigonometric Functions: Mastering [(𝒕𝒂𝒏 𝒙+ 𝒄𝒐𝒕 𝒙)/(𝟑𝒆^𝒙)] with the Quotient Rule

Get ready to enhance your calculus skills by learning how to differentiate the complex expression [(𝒕𝒂𝒏 𝒙+ 𝒄𝒐𝒕 𝒙)/(𝟑𝒆^𝒙)]  using the powerful quotient rule. This tutorial is perfect for students tackling advanced calculus topics, offering clear, step-by-step guidance to help you understand the intricacies of trigonometric derivatives.
#Calculus #QuotientRule #Trigonometry #MathSkills #Differentiation


let's solve this calculus problem step

by step we need to find the derivative

of x to the 4th power time e to the X

this is a product of two functions so

we'll need to use the product rule the

product rule states that the derivative

of a product U * v = u * the derivative

of V plus v * the derivative of U in our

case u = x 4th power and v = e to X

let's apply the product rule first we

write X 4th power time the derivative of

e to X Plus e x * the derivative of x 4

power remember the derivative of e to X

is just e to X and the derivative of x

to the 4th power is 4 x Cub now let's

factor out the common terms

both terms have e to X so we can factor

that out when we do we get e to x * the

X to 4 + 4x Cub finally we can factor

out X cubed from both terms inside the

parentheses this gives us X cubed e to

the x * the quantity X+ 4 which is our

final

answer let's visualize what's happening

with this derivative here's the original

function X to 4th * e to X in blue and

here's its derivative in red notice how

the derivative grows even faster than

the original function due to both the

polom and exponential terms to summarize

we use the product rule to find this

derivative carefully differentiated each

part and then factored our result to get

it into its simplest form the

visualization helped us see how this

function and its derivative behave

Finding the Derivative of 𝑥^4𝑒^𝑥  Using the Product Rule | Step-by-Step Explanation

In this video, learn how to find the derivative of 𝑥^4𝑒^𝑥
 using the Product Rule in calculus. 🎓
📌 What You'll Learn in This Video:
What is the Product Rule?
Step-by-step solution to differentiate 𝑥^4𝑒^𝑥
Formula breakdown and example problem-solving.
This example is perfect for beginners in calculus or anyone reviewing the Product Rule for differentiation!
💡 Don't forget to LIKE, SHARE, and SUBSCRIBE for more math tutorials!
#Calculus #ProductRule #Differentiation #Derivative #MathTutorial #LearnMath #MathEducation #MathHelp


let's solve this derivative step by step

using the product rule we need to find

the derivative of e to x * sinx first

let's recall the product rule the

derivative of a product of two functions

U * V equal U * the derivative of V plus

v * the derivative of U let's identify

our functions in this case u = e to X

and V = sin x now let let's find the

derivatives of each function separately

the derivative of e to X is e to X and

the derivative of sinx is cine X let's

apply the product rule using these

pieces we start with E to x * the

derivative of sin x which is cine X then

we add sin x * the derivative of e to X

which is e to X finally we can factor

out e to X since it appears in both

terms this gives us our final answer e

to x * the quantity cosine X Plus sinx

let's verify this makes sense we used

the product rule correctly found the

derivatives of both functions and

factored out the common term our answer

shows that the derivative of e to the X

sinx involves both trigonometric

functions and the exponential function

which we would expect remember when

dealing with products of functions

especially when one is exponential and

the other

trigonometric the product rule is our go

to Method this example shows how

powerful and straightforward the product

rule can be

Finding the Derivative of e^x sin(x)  Using the Product Rule | Step-by-Step Explanation

In this video, discover how to differentiate e^x sin(x)) using the Product Rule in calculus.

What You'll Learn:
- Definition of the Product Rule
- Step-by-step differentiation of e^x sin(x)
- Formula breakdown and example problem-solving

This example is ideal for calculus beginners or anyone reviewing the Product Rule for differentiation!

💡 Don't forget to LIKE, SHARE, and SUBSCRIBE for more math tutorials!
#Calculus #ProductRule #Differentiation #Derivative #MathTutorial #LearnMath #MathEducation #MathHelp


welcome today we'll solve a challenging

derivative problem using logarithmic

differentiation we need to find the

derivative of y = theare < TK of thean x

- 1 * x - 2 / the quantity x - 3 * x - 4

this looks complicated to differentiate

directly instead we'll use logarithmic

differentiation the strategy is to take

the the natural log of both sides first

then differentiate let's start by taking

the natural log of both sides when we do

this we get Ln of y equal Ln of our

original expression now we can use the

properties of logarithms to break this

down first we can separate the square

root into its numerator and denominator

then we can split this into the

difference of logarithms using using the

properties of logarithms again we can

break down the square roots this gives

us 12 * the natural log of each term now

we can differentiate both sides with

respect to X on the left side we get 1/

y *

dydx on the right side we use the chain

rule to differentiate each logarithm

term to solve for dydx we multiply both

sides by y remember that y Y is our

original function finally we substitute

back the original expression for y to

get our complete derivative and there we

have it we've found the derivative using

logarithmic

differentiation this method is

particularly useful when dealing with

products quotients and Roots as it often

simplifies the differentiation process

Mastering Derivatives with Logarithmic Differentiation [√((𝑥−1)(𝑥−2)/(𝑥−3)(𝑥−4) )]

Question:
Find the derivative of the function y= [√((𝑥−1)(𝑥−2)/(𝑥−3)(𝑥−4) )]  using logarithmic differentiation.

In this video, we explore the technique of logarithmic differentiation to find the derivative of complex functions. Follow along as we break down each step clearly, providing both mathematical steps and explanations. Perfect for calculus students looking to deepen their understanding of derivatives!
#Calculus #Derivatives #MathTutorial #LogarithmicDifferentiation #MathTips


today we'll find the derivative of

secant x using the first principes

definition of derivatives let's start

with our function let F ofx = secant x

by definition the derivative is the

limit as H approaches Z of f ofx + hus F

ofx all divided H substituting our

secant function we get the limit as H

approaches Zer of secant

of x + h- secant x all divid H remember

that secant is 1 / cosine so we can

rewrite this as the limit as H

approaches 0 of 1 / cosine of x + H - 1

/ cosine X all divid H to subtract these

fractions we need a common denominator

multiplying both terms by cosine x

cosine of X Plus h/ cosine x cosine of x

+ H gives us cosine x - cosine of x + H

over cosine x cosine of x + H now comes

the key step we can use the cosine

difference formula which states that

cosine a minus cosine B = -2 * sin of a

plus b/ 2 * sin of a minus B /2 let's

simplify the arguments the first sign

term becomes s of x + h / 2 and the

second becomes s of H /2 now we can

split this into separate limits the

limit of s of x + h / 2 as H approaches

0 is sin x the limit of s of h / 2 over

h / 2 as H approaches 0 is 1 and the

limit of cosine of x + H as H approaches

0 is cosine X finally we can rewrite

this as sin x cosine x * 1 / cosine X

which is tangent x * secant x therefore

the derivative of secant x equal tangent

x * secant x to visualize this result

let's look at the graphs of SEC X in

blue and its derivative tangent x * SEC

X in red

Understanding the Derivative of secx  Using First Principles

Master the fundamentals of calculus by deriving the exponential function sec3x from first principles. Explore our step-by-step explanations and examples to enhance your grasp of derivatives and exponential functions.
#Calculus #Derivatives #ExponentialFunctions #MathTutorial #FirstPrinciples #tan3x


let's solve this complex derivative

problem step by step we'll start with

our original function the first step is

to convert both terms to exponential

form using the property that a to the

power be equals e to the power of B *

natural log of a for the first term x to

the power of sin x becomes E power of

sin x * natural log of x similarly for

the second term sin x to the power of X

becomes E power of x * natural log of

sinx now we can find the derivative

using the chain rule remember when

differentiating e to some power we get e

to that same Power Times the derivative

of the power let's handle the first term

we need to find the derivative of sinx *

natural log X using the product rule

this gives us s x * 1 /x + cosine x *

natural log X for the second term we

need to differentiate x * natural log of

sin x this gives us natural log of sin x

+ x times the derivative of natural log

of sinx which is coent X finally we can

convert back to our original form by

replacing e to the power of sin x Ln x

with x the power of sin x and e to the

power of X Ln sin x with sin x the power

of X this gives us our final answer this

problem demonstrates several important

calculus Concepts converting exponential

expressions using the chain Rule and

product rule and handling complex

derivatives involving trigonometric and

logarithmic functions

Complex Derivatives: Step-by-Step Solution of y = x^(sin x) + (sin x)^x | Calculus Made Easy

Question:
Find the derivative of y = x^(sin x) + (sin x)^x

🔍 In this detailed calculus tutorial, we break down the solution to a challenging derivative problem involving variable exponents. Learn how to:
• Convert complex expressions using natural logarithms
• Apply the chain rule and product rule effectively
• Handle logarithmic and trigonometric derivatives
• Transform expressions back to their original form
Perfect for:
📚 Calculus students
🎓 Mathematics majors
📝 Exam preparation
🧮 Advanced differentiation practice
#Calculus #Mathematics #ChainRule #Derivatives #MathTutorial #ExamPrep #CalculusTutorial #MathEducation


let's solve this calculus problem step

by step using the quotient rule we need

to find the derivative of the function

1us tangent thet / 1 + tangent Theta

let's recall the quotient rule the

derivative of U over V equal V * the

derivative of U minus uu * the

derivative of V all / v² Let's identify

our u and v in this case U equal 1 -

Tang thet and V = 1 + tangent thet next

we need to find the derivatives of u and

v the derivative of U is secant 2 Theta

and the derivative of V is secant 2

Theta now let's substitute everything

into the quotient rule formula we get 1

+ tangent Theta * secant 2 thet - 1 -

Tang thet * secant 2 Theta all over 1 +

tangent thet 2 let's expand the

numerator first term gives us secant 2

Theta -^ 2et * tangent Theta second term

gives us secant 2 th +^ 2 thet * tangent

thet now let's combine like terms the

tangent Theta terms cancel out and we're

left with negative 2 secant 2 Theta in

the numerator and this is our final

answer we've successfully used the

quotient rule to find the derivative of

this function to verify this is correct

we can check that one our answer is a

quotient two we correctly applied the

quotient rule formula and three we

properly simplified our expression

Quotient Rule Masterclass: Differentiating Complex Fractions [(𝟏−𝒕𝒂𝒏 𝜽)/(𝟏+𝒕𝒂𝒏 𝜽)] with Ease

Dive into the world of calculus with this in-depth look at using the quotient rule for complex fractions. We’ll expertly guide you through differentiating the function [(𝟏−𝒕𝒂𝒏 𝜽)/(𝟏+𝒕𝒂𝒏 𝜽)]  , helping you master this essential calculus technique. Perfect for students wanting to strengthen their differentiation skills.
#Calculus #Differentiation #QuotientRule #MathEducation #LearnMath


let's solve this calculus problem step

by step we need to differentiate the

function Ln of sin x^2 with respect to X

let's start by applying the chain rule

remember the derivative of natural log

of U is 1/ U * the derivative of U so in

our case U is sin x^2 now we need to

find the derivative of sin x^2 the

derivative of s of V is cosine V time

the derivative of V here V is X2 let's

rearrange this expression we can combine

1 / sin x^2 and cine x^2 into cine x^2

over sin x^2 now for the last part we

need to differentiate x^2 the derivative

of x^2 is 2x multiply this by our

previous expression for our final step

recall that cosine over s is the

definition of coent therefore our final

answer is 2x * cent of x^2 and that's

how we differentiate Ln of sin x^2 we

use the chain rule twice once for the

natural log and once for the sign

function and then simplified our result

using the definition of cotangent

Mastering Derivatives Calculating the Derivative of {lnsinx²}

Join us as we break down the process of finding the derivative of {lnsinx²}  step-by-step. Learn how to apply the chain rule effectively and understand the use of trigonometric identities. Perfect for students tackling calculus and differentiation!
#Calculus #Derivatives #MathTutorial #Trigonometry #Differentiation


let's solve this challenging

trigonometric derivative problem step by

step we need to find the derivative of

sin inverse of 2 a x * < TK of 1 - A 2

x^2 let's start by letting y equal our

expression then we'll make a clever

substitution to simplify this let's say

ax equal sin Theta this substitution

allows us to rewrite our expression

inside the sin inverse

we now have 2 sin thet * the < TK of 1 -

sin^2 now we can use a fundamental

trigonometric identity 1 - sin^2 thet =

cine 2 thet the square root of cosine 2

thet is simply cosine Theta and 2 sin

thet * cosine Theta is the double angle

formula for sin 2 Theta when we have S

inverse of s of an angle they cancel out

leaving us with Just 2 Theta and since

Theta was sin inverse of ax our final

expression for y is 2 sin inverse of ax

now we can differentiate this expression

the derivative of y with respect 2x = 2

* the derivative of sin inverse of ax

using the chain Rule and the derivative

formula for sin inverse we get 2 * 1 <

TK of 1 - -

ax^2 time the derivative of ax our final

answer is 2 over the < TK of 1 - A 2

x^2 this problem demonstrates the power

of clever substitution and trigonometric

identities in simplifying complex

derivatives

Understanding the Differentiation of [𝑠𝑖𝑛^(−1){2𝑎𝑥 √(1−𝑎^2 𝑥^2 )}] Step-by-Step Simplification

Question
Find Derivative of  [𝑠𝑖𝑛^(−1){2𝑎𝑥 √(1−𝑎^2 𝑥^2 )}].

In this video, we delve into the calculus behind differentiating the complex expression  [𝑠𝑖𝑛^(−1){2𝑎𝑥 √(1−𝑎^2 𝑥^2 )}]. Follow along as we simplify and differentiate step-by-step, using trigonometric identities and the chain rule. Whether you’re a calculus student or just a math enthusiast, this detailed explanation will help you grasp this intriguing problem. Perfect for those preparing for exams or seeking a deeper understanding of calculus!
#Calculus #Differentiation #MathTutorial #InverseTrigonometry #ChainRule #MathHelp #Education


let's solve this complex derivative step

by step we're going to find the

derivative of the square root of tangent

of e to x^2 first let's rewrite this

expression the square root can be

written as a power of 12 so our

expression becomes the derivative of

tangent of e to x^2 all rais to the 1/2

power now we can apply the chain rule

remember when we have a function raised

to a power we multiply by the power

reduce the power by one and multiply by

the derivative of the inner function

this gives us 12 * tangent of e x^2 to

the2 Power Times the derivative of the

inner function for the derivative of

tangent we use the fact that the

derivative of tangent is secant squar so

we get 12 * 1 root of tangent of e to

x^2 * secant 2 of e to x^2 * the

derivative of e to x^2 now we need to

find the derivative of e to x^2 using

the chain rule again this gives us e to

x^2 * 2x so our expression becomes 12 *

secant 2 of e to x^2 over the square

root of tangent of e to x^2 * e x^2 * 2x

finally we can simplify this expression

the 1/2 and two cancel out giving us our

final answer x * e to x^ 2 * secant 2 of

e to x^2 all over the square < TK of

tangent of e to x^2 let's review the key

steps we used first we rewrote the

square root as a power of 1/2 then we

applied the chain rule to handle the

outer function next we used the fact

that the derivative of tangent is secant

squared finally we found the derivative

of the exponential term and simplified

our expression this problem shows how we

can break down complex derivatives into

simpler steps using the chain Rule and

our knowledge of basic derivatives

9.3-1.7 How to Differentiate (√𝒕𝒂𝒏𝒆^𝒙^𝟐)  Step-by-Step

Unlock the secrets of calculus with this detailed, step-by-step guide to differentiating (√𝒕𝒂𝒏𝒆^𝒙^𝟐) . We’ll break down the process using the chain rule and power rule, making complex derivatives easy to understand. Perfect for students and anyone looking to strengthen their calculus skills!
#Calculus #Differentiation #ChainRule #MathTutorial #MathHelp #cpmrana


let's solve the step by step we need to

find the derivative of cine 2 of x^2

first let's understand what we're

dealing with this is a composite

function we have cosine squar and inside

that we have X2 we'll need to use the

chain rule here let's apply the chain

rule the chain rule tells us that we

multiply the derivative of the outer

function with respect to U times the

derivative of the inner function with

respect to X let's find the derivative

of the outer function first to

differentiate cosine squar we use the

power rule the power comes down in front

and we multiply by the derivative of

what's inside so the derivative of

cosine 2 of U is 2 cosine of U * the

derivative of cosine of U which is

negative s of U now for the inner

function the derivative of x^2 is simply

2X now let's put it all together we

multiply -2 cosine of u s of U where U

is x^2 * 2x this gives us -2 cosine of

x^2 sin of x^2 * 2x let's simplify this

expression we can rearrange the terms to

get -2X * 2 sin of x^2 cosine of x^2

finally we can use the double angle

formula remember remember that 2 sin of

x cosine of x = sin of 2x so our final

answer is -2X sin of

2x^2 and there we have it we've found

the derivative using the chain Rule and

the double angle formula let's review

the key steps we use the chain rule to

break down the derivative found the

derivatives of both the outer and inner

functions multipli them together and

simplified using the double angle

formula

9.3-1.9 Mastering Derivatives A Step by Step Guide to (𝑐𝑜𝑠^2^𝑥^2)

In this video, we break down the process of differentiating the function (𝑐𝑜𝑠^2^𝑥^2)  step by step. Learn how to apply the chain rule, simplify expressions, and utilize trigonometric identities to find the derivative. Perfect for students and math enthusiasts looking to deepen their understanding of calculus concepts!
#Calculus #Differentiation #ChainRule #MathTutorial #Trigonometry


let's solve this challenging

differentiation problem step by step we

need to differentiate the function f ofx

equals the natural log of e to x * the <

TK of x - 1 overx + 1 let's break this

down into manageable steps using the

properties of logarithms first using the

property that the log of a product

equals the sum of logs we can split this

into the natural log of e to X plus the

natural log of the square root term next

we can use the power property of

logarithms to bring down the 1/2 power

this gives us natural log of e to X

+2 * the natural log of x -1 / x +

1 now we can use another property of

logarithm to split the fraction this

gives us natural log of e to x + 12 *

the natural log of x

-1 - 12 * the natural log of x +

1 now we can apply the chain rule to

differentiate each term remember when

differentiating natural log we get one

over the function time the derivative of

the

function let's evaluate each derivative

the derivative of e to X is e to X and

the derivative of X+ or minus one is

just

one now we can simplify e to X over e to

x =

1 to combine these fractions we need a

common denominator we'll multiply

everything by 2 X1 x +1 over 2 X1 x +

1 Let's expand the numerator in the

first term we get 2 x^2 - 2 then we add

x +1 and subtract x

-1 this simplifies to 2x^2 in the

numerator and our final answer is x^2

over x^2 -

1 this problem demonstrated several

important calculus Concepts the chain

rule logarithm properties and fraction

manipulation remember to break down

complex problems into small manageable

steps

Mastering Logarithmic Differentiation [𝒍𝒏 {𝒆^𝒙 ((𝒙−𝟏)/(𝒙+𝟏))^(𝟏/𝟐) }] Step-by-Step Guide

Question:
Differentiate the expression [𝒍𝒏 {𝒆^𝒙 ((𝒙−𝟏)/(𝒙+𝟏))^(𝟏/𝟐) }] with respect to x.

Discover how to differentiate complex logarithmic expressions with ease! In this video, we’ll break down the process of differentiating [𝒍𝒏 {𝒆^𝒙 ((𝒙−𝟏)/(𝒙+𝟏))^(𝟏/𝟐) }] step by step. Perfect for students and anyone looking to strengthen their calculus skills. Follow along and master the art of logarithmic differentiation with detailed explanations. Don’t forget to like, subscribe, and hit the bell icon for more math tutorials!
#Calculus #Differentiation #MathTutorial #Logarithms #STEM #Education


let's solve this differentiation problem

step by step we need to find the

derivative of e to the x * tangent s of

X first let's identify what we're

dealing with this is a product of two

functions e to X and tangent s of X when

we have a product we need to use the

product rule let's apply the product

rule the first term will be e to x * the

derivative of tangent 2 x

plus t^ 2 x * the derivative of e to X

now let's handle each term separately

for the first term we need to use the

chain rule on tangent 2 x this gives us

2 * tangent x times the derivative of

tangent X for the second term we know

the derivative of e to the x is a Time e

to X let's continue simplifying the

derivative of tangent X is secant 2 x

and the derivative of x is simply a now

let's combine like terms we can factor

out e to X from both terms then we can

factor out tangent X from what remains

and there we have our final answer the

derivative of e to x * tangent 2 x is e

to x * tangent x * the quantity 2^ 2 X

plus a tangent X X this problem showed

us how to use both the product rule and

chain rule together to handle complex

derivatives

Mastering Differentiation [ 𝑒^𝑎𝑥 𝑡𝑎𝑛^2  𝑥] Solving  Step-by-Step

Question: Differentiate [ 𝑒^𝑎𝑥 𝑡𝑎𝑛^2  𝑥]   with respect to x.
In this video, we explore how to differentiate the function [ 𝑒^𝑎𝑥 𝑡𝑎𝑛^2  𝑥] using the product and chain rules. Follow along with clear, step-by-step explanations to enhance your calculus skills.
#Calculus #Differentiation #MathTutorial #MathHelp #ChainRule #ProductRule #MathTips


let's solve this derivative step by step

we're going to find the derivative of 1

over the cube root of 4 - 3x first let's

rewrite this expression in a form that's

easier to work with we can rewrite the

cube root in the denominator as a power

of - 13 this gives us the derivative of

4 - 3x to the - 13 power now we can use

the power rule and change rule together

the power rule tells us that when we

differentiate x to the N power we

multiply by the power and reduce the

exponent by one and since we have an

inner function of 4 - 3x we'll also need

to multiply by its derivative let's

apply these rules first we multiply by -

13 then we decrease the exponent by one

making it

-4/3 then we multiply by the derivative

of the inner function which is -3 now

let's simplify the negative signs cancel

out and multiplying - 13 by -3 gives us

POS 1 this leaves us with 4 - 3x -43

power finally we can write this as a

fraction giving us 1 over 4 - 3x^ 43

power this is our final answer let's

review the key steps we took first we

rewrote the cube root as a negative

power then we applied the power rule and

chain rule together and finally we

simplified our expression to get our

answer

9.3-1.2 Mastering Derivatives Simplifying Complex Expressions Step by Step-(1/∛(4−3𝑥))

In this video, we break down the process of finding the derivative of a complex expression involving cube roots. Learn how to apply the power and chain rules, and simplify your answers with ease. Follow along as we provide a detailed, step-by-step explanation that will enhance your calculus skills. Perfect for students and math enthusiasts looking to deepen their understanding of derivatives!
#Calculus #Derivatives #MathTutorial #ChainRule #PowerRule #MathHelp #Education


let's find the derivative of natural

logarithm using first principles we

start with f of x = natural log of x

step one let's Define f ofx + H when we

replace x with x + H we get f of x + H =

natural log of x + H step two we'll use

the definition of derivative the

derivative is the limit as h appr

apprach is z of f of x + hus F ofx all

divided H substituting our Expressions

we get the limit as H approaches zero of

natural log of x + hus natural log of x

all divided by

H step three now we'll use the

properties of logarithms the difference

of logarithms equals the logarithm of

their quotient so this becomes the limit

as H approaches 0 of 1H * the natural

log of x + h / X step four let's

simplify the fraction inside the

logarithm we can rewrite x +

HX as 1 +

HX step five now comes the key step

we'll use the series expansion for

natural log of 1 + y natural log of 1 +

y = y - y^2 / 2 + plus y Cub over 3us

and so on when we substitute y = HX we

get this expression step six let's

simplify by collecting terms with h

after rearranging we get 1X - h / 2x^2 +

h^ 2 over 3x cub and so on finally as H

approaches zero all terms contain

meaning H will vanish therefore the

derivative of natural log of x = 1 /x

Let's visualize this result here's the

graph of y equals natural log of x and

at any point its derivative or slope of

the tangent line is 1 /x at that point

Understanding the Derivative of lnx  Using First Principles

Dive into the fundamentals of calculus as we explore how to determine the derivative of the exponential function lnx using first principles. Follow along with step-by-step explanations and examples to enhance your understanding of derivatives and exponential functions.
#Calculus #Derivatives #ExponentialFunctions #MathTutorial #FirstPrinciples #lnx


welcome today we'll learn how to find

the derivative of a complex function

using logarithmic differentiation our

function is y = 1 + x^2 raed to^ of 2x

this is a complex function because it

has both a variable base and a variable

exponent the base is 1 + x^2 and the

exponent is 2x when we see this kind of

structure logarithmic differentiation is

often the best approach let's start by

taking the natural logarithm of both

sides this gives us Ln of Y = Ln of 1 +

x^2 raed to 2x now we can use the

logarithm property that states the

natural log of a power equals the

exponent times the natural log of the

base this gives us Ln of y = 2x * Ln of

1 + x^2 now we can differentiate both

sides with respect to X on the left side

we get 1/ y *

dydx this comes from the chain rule for

the right side we need to use the

product rule since we have 2X * Ln of 1

+ x^2 the product rule states that the

derivative of U * v = u Prime * V + U *

V Prime let's apply this the derivative

of 2x is just 2 and we multiply that by

Ln of 1 + x^2 then we add 2x * the

derivative of Ln of 1 + x^2 for this

last part we use the chain rule the

derivative of Ln of 1 + x^2 is 1 1 + x^2

* the derivative of what's inside which

is 2x now we can solve for dydx by mul

multiplying both sides by y remember

that Y is our original function 1 + x^ 2

to 2X and this is our final answer let's

break down what we did first we took the

natural log of both sides then we used

logarithm properties to simplify next we

differentiated both sides using the

chain rule on the left and the product

rule on the right finally we solve for

dydx by multiplying both sides by y

logarithmic differentiation is a

powerful technique that helps us handle

complex functions especially those with

variable bases and exponents remember to

use it whenever you see a function that

would be difficult to differentiate

using standard rules

Mastering Logarithmic Differentiation: Solving (1 + x²)^(2x) Step-by-Step

Question: Find the derivative of y = (1 + x²)^(2x) using logarithmic differentiation.

📚 In this comprehensive calculus tutorial, we break down the solution to finding the derivative of (1 + x²)^(2x) using logarithmic differentiation. Perfect for calculus students and math enthusiasts!
🎯 What you’ll learn:
• Why logarithmic differentiation is useful
• How to handle functions with variable bases and exponents
• Step-by-step application of logarithm properties
• Using product rule and chain rule effectively
• Simplification techniques for complex derivatives
✏ Prerequisites:
• Basic differentiation rules
• Understanding of natural logarithms
• Knowledge of product and chain rules
#Calculus #Mathematics #LogarithmicDifferentiation #MathTutorial #CalcTutorial #MathEducation #LearnMath #Derivatives


let's solve this problem step by step we

need to find all natural numbers X where

x^2 is greater than 9 and X Cub is less

than

130 let's create a table to check

different values of X we'll start with

xal 1 and work our way up until we find

all possible

solutions let's check x = 1 1^ 2ar is 1

which is not greater than 9 and 1 cubed

is also 1 Which is less than

130 but since 1 2ar isn't greater than 9

this isn't a solution 4X = 2 2^2 is 4

still not greater than 9 2 cubed is 8

Which is less than 130 but again 2^ 2ar

isn't greater than 9 so this isn't a

solution either 4X = 3 3^2 is is 9 which

is equal to 9 but not greater than 9 3

cubed is 27 Which is less than 130 but

3^2 still isn't greater than 9 now 4X =

4 4^ 2ar is 16 which is greater than 9

and 4 cubed is 64 Which is less than

130 this means x = 4 is a solution let's

check X = 5 5^ 2ar is 25 which is

greater than 9 and 5 cubed is 125 which

is still less than

130 so x = 5 is also a solution finally

let's check x = 6 6^ 2 is 36 which is

greater than 9 but 6 cubed is

26 which is greater than

130 so x = 6 is not a solution let's

visualize this on a number line to see

our solution set more

clearly therefore our solution set is

the set containing four and five we can

write this in set notation as 4 five

these are the only natural numbers that

satisfy both conditions having a square

greater than 9 and a cube less than 130

Solving Set Problems with Natural Numbers {𝑥^2 grater than 9 and  𝑥^3 less than130}.

Dive into solving set theory problems involving natural numbers! In this video, we explore a specific problem to find the subset of natural numbers x where {𝑥^2 grater than 9 and  𝑥^3 less than130}. Follow along as we meticulously work through each condition to identify the solution. Ideal for understanding mathematical expressions and their applications in set problems.
#SetTheory #NaturalNumbers #MathProblems #Algebra #MathTutorial #ProblemSolving


let's solve this calculus problem step

by step using the quotient rule we need

to find the derivative of 1 + sinx / 1 -

sinx first let's recall the quotient

rule formula for a fraction U over V the

derivative is V *

DX minus U *

dvdx all over V ^2 let's identify our u

and v functions here u = 1 + sinx and V

= 1 - sinx now let's find the

derivatives of u and v the derivative of

U is cine X and the derivative of V is

cosine X let's substitute these into the

quotient rule formula we get 1 - sin x *

cosine X - 1 + sin x * cosin X all over

1 - sin x^2 now let's expand the

numerator from the first term we get

cine x - sin x cosine X from the second

term we get plus cosine X Plus sin x

cine X let's simplify the numerator the

sin x cine X terms cancel out and we're

left with cine X+ cosine X which is 2

cosine X and there we have our final

answer the derivative of 1 + sin x over

1 1 - sin x = 2 cosin x / 1 - sin x^2

this problem demonstrates how to use the

quotient rule to find the derivative of

a complex fraction involving

trigonometric functions remember to

carefully identify your u and v

functions find their derivatives and

then apply the quotient rule formula

Differentiating [(𝟏+𝒔𝒊𝒏𝒙)/(𝟏-sinx)] Using the Quotient Rule: Step-by-Step Tutorial

Master the art of differentiation with this detailed tutorial on finding the derivative of the function [(𝟏+𝒔𝒊𝒏𝒙)/(𝟏-sinx)] using the quotient rule. This video walks you through each step and simplifies complex calculus concepts for easy understanding. Perfect for students, educators, and math enthusiasts aiming to strengthen their calculus skills.
#Calculus #Differentiation #QuotientRule #MathTutorial #StepByStep


welcome today we'll solve a challenging

derivative problem using logarithmic

differentiation let's find the

derivative of y = sin of X ra to the

power of tangent X this is a complex

function because we have a trigonometric

function raised to another trigonometric

function the best approach here is to

use logarithmic differentiation let's

start by taking the natural log of both

sides using the logarithm property that

says log of a to the power Bal B * log

of a we can simplify this to Ln y =

tangent x * Ln of sin x now we'll

differentiate both sides with respect to

X on the left side we get 1/ y *

dydx on the right side we'll need to use

the product rule since we have the

product of two functions let's evaluate

each derivative separately for Ln of

sinx the derivative is cine X sin x and

for tangent X the derivative is secant 2

x now let's solve for

dydx first multiply both sides by y

let's simplify the trigonometric terms

notice that tangent x * cine x/ sin x

simplifies to just one since cos sin x

over sin x is 1 / tangent X finally we

can substitute back the original

function for y giving us our final

answer the derivative equals sinx to the

power of tangent X time the quantity 1 +

natural log of sin x * SEC 2 x let's

review the key steps we used first we

took the natural log of both sides then

we used logarithm properties to simplify

we differentiated using the product rule

and chain rule finally we substituted

back the original function this is a

great example of how logarithmic

differentiation can help us find

derivatives of complex exponential

functions

Finding the Derivative of (sin x)^(tan x) | Logarithmic Differentiation Made Easy

Question: Find the derivative of y = (sin x)^(tan x) using logarithmic differentiation.

🔍 Master logarithmic differentiation with this step-by-step guide to finding the derivative of (sin x)^(tan x). Perfect for Calculus students and math enthusiasts!
In this video, we cover:
• Logarithmic differentiation technique
• Product rule application
• Chain rule implementation
• Trigonometric derivatives
• Step-by-step solution walkthrough
🎯 Learning Outcomes:
Understand when to use logarithmic differentiation
Master complex derivative problems
Apply multiple calculus rules together
Simplify trigonometric expressions
📚 Prerequisites:
Basic differentiation rules
Knowledge of trigonometric functions
Understanding of logarithm properties
#Calculus #Mathematics #MathTutorial #CalcTutorial #MathEducation #LogarithmicDifferentiation #LearnMath #MathHelp #STEM


let's solve a challenging

differentiation problem using

logarithmic differentiation we need to

find the derivative of y = e to x^2 *

inverse tangent of X all divided by the

square root of 1 + x^2 this is a complex

function involving products quotients

and compositions the best approach is to

take the natural logarithm of both sides

this gives us l n of Y = Ln of our

entire expression using the properties

of logarithms we can split this into the

difference of logarithms the numerator

becomes Ln of e x^2 * inverse tangent of

xus Ln of the < TK of 1 + x^2 again

using logarithm properties we can split

the first term into Ln of e to X 2 + Ln

of inverse tangent of X and remember Ln

of a square T is 12 Ln of the expression

now we can differentiate both sides with

respect to X on the left side we get 1 /

y *

dydx on the right side we use the chain

rule for each term the derivative of x^2

is 2x for the natural log terms we use

use the chain rule this gives us 1 /

inverse tangent of x * its derivative

which is 1 1 + x^2 and for the last term

we get -2 * 1 1 + x^2 * 2x now we can

solve for dydx by multiplying both sides

by Y and finally we substitute back the

original expression for y to get our

complete derivative this gives us e to

x^ 2 * inverse tangent of X over < TK of

1 + x^2 times the expression in Brackets

and that's how we use logarithmic

differentiation to find the derivative

of this complex function the key steps

were taking the natural log of both

sides using log properties to simplify

and then differentiating term by term

Differentiating Complex Logarithmic Functions [(𝑒^(𝑥^2 ) 𝑡𝑎𝑛^(−1) 𝑥)/√(1+𝑥^2 )] Step-by-Step

Problem
Differentiate:  [(𝑒^(𝑥^2 ) 𝑡𝑎𝑛^(−1) 𝑥)/√(1+𝑥^2 )]

Unlock the secrets of differentiating complex logarithmic functions with this easy-to-follow, step-by-step guide. We’ll break down each part, making it simple to tackle even the toughest math problems!
#MathTutorial #Differentiation #LogarithmicFunctions #Calculus #StepByStep


let's solve a challenging derivative

problem using logarithmic

differentiation our goal is to find the

derivative of y = x raed to the natural

log of x this is a complex function

where X appears both in the base and in

the exponent let's start by letting y

equal x to the lnx power the key to

solving this type of problem is to take

the natural logarithm of both sides

using the logarithm property that says

Ln of X raised to a power equals that

power * lnx we can simplify this to Ln x

* Ln X or Ln x^2 now comes the key step

we're going to differentiate both sides

with respect to X on the left side we

get 1/ y *

dydx and on the right side we get 2 Ln x

* the derivative of Ln X which is 1 /x

Let's solve for dydx by multiplying both

sides by y then we can substitute back y

= x to the Ln X power finally we can

simplify this to get our answer let's

highlight the key techniques we used to

solve this problem first we used

logarithmic differentiation by taking

the natural log of both sides second we

used logarithm properties to simplify

the expression and finally we use the

chain Rule and product rule for

differentiation therefore the derivative

of x to the Ln X power is 2 * x^ of Ln X

- 1 * lnx this is a great example of how

logarithmic differentiation can help us

solve complex derivative problems

Logarithmic Differentiation: Solving [y = x^(ln x)] | Calculus Made Clear

Question:
Find the derivative of [y = x^(ln x)] with respect to x.

Master logarithmic differentiation with this step-by-step solution of y = x^(ln x). In this comprehensive tutorial, we break down a challenging calculus problem that combines exponential and logarithmic functions.
🎯 What you’ll learn:
• Why logarithmic differentiation is necessary
• How to apply natural logarithm properties
• Step-by-step differentiation process
• Common mistakes to avoid
• Simplification techniques
🔍 Perfect for:
• Calculus students
• AP Calculus preparation
• College math review
• IB Mathematics
💡 Free practice problems in description!
#Mathematics #Calculus #MathTutorial #Education #LearnMath #APCalculus #CollegeMath #StudyWithMe #MathHelp #CalculusTricks


let's solve this complex derivative

problem step by step our goal is to find

the derivative of this expression with

respect to X let's break this down into

manageable steps first let's set our

expression equal to Y to make it easier

to work with now let's expand the

numerator when we Square x^2 + 1 we get

x 4 + 2x^2 + 1 in the denominator we can

re write the cube root as x^ the 2/3

power now comes a key step we can

distribute the division by x to the 2/3

to each term in the numerator when we do

this we subtract the exponents let's

simplify these exponents for - 2/3

becomes 10/3 2 - 2/3 becomes 4/3 and

-2/3 stays as is now we can

differentiate each term using the power

rule remember when using the power rule

we multiply by the power and reduce the

exponent by one let's apply the power

rule to each term for the first term we

multiply by 10/3 and subtract one from

the exponent for the second term we

multiply by 4/3 and subtract one for the

last term we multiply by -2/3 and

subtract 1 1 now let's simplify these

exponents 10/3 - 1 becomes 7/3 4/3 - 1

becomes 1/3 and -2/3 - 1 becomes -

5/3 finally we can factor out 2/3 to

write our answer in a more compact form

our final answer is 2/3 times the

expression in Brackets and there we have

it we've successfully found the

derivative of this complex expression by

breaking it down into manageable steps

and using the power rule

Mastering Complex Derivatives [(𝑥^2+1)^2/∛(𝑥^2 )]  Solving Power Functions with Cube Roots |

Question: Find the derivative of [(𝑥^2+1)^2/∛(𝑥^2 )]    with respect to x.

Master the art of solving complex derivatives involving power functions and cube roots in this detailed tutorial. We break down the solution of d/dx[(x²+1)²/∛(x²)] step by step.
In this video, you’ll learn:
📚 How to simplify complex fractions with roots
✏ Converting cube roots to fractional exponents
🔢 Laws of exponents and their applications
📐 Power rule for differentiation
💡 Step-by-step algebraic simplification
Perfect for:
• Calculus students
• Math majors
• Engineering students
• Test preparation (AP Calculus, IB Mathematics)
#Mathematics #Calculus #Derivatives #MathTutorial #PowerRule #StudyWithMe #ExamPrep #MathEducation #STEM #EngineeringMath


let's solve this complex inverse cosine

differentiation problem step by step

first let's look at what we're trying to

find to solve this we'll use a

substitution trick let's say x = cine

Theta this means Theta equal inverse

cosine of x now let's substitute this

into our original equation we get y =

inverse cosine of < TK 1 + Co cosine

Theta / 2 here's where we use a key

trigonometric identity 1 + cine thet = 2

cosine 2 Theta / 2 using this identity

our expression becomes inverse cosine of

cosine 2 Theta over 22 power simplifying

the fraction inside we get inverse

cosine of cosine 2 Theta /

2^2 power taking the square root we get

inverse cosine of cosine Theta / 2 now

here's the key Insight inverse cosine

and cosine are inverse functions so they

cancel out leaving us with just Theta

over 2 remember that Theta equal inverse

cosine X so y = 12 inverse cine X now we

can differentiate the derivative of y

with respect to x equals the derivative

of 12 inverse cine X we can pull out the

1/2 and then use the fact that the

derivative of inverse cosine X

is-1 over < TK of 1 - x^2 therefore our

final answer is -1/2 * the < TK of 1 -

x^2 and that completes our solution the

key was using the substitution xal cine

Theta

and applying the double angle formula

for cosine

Understanding Inverse Trigonometric Functions [𝑐𝑜𝑠^(−1)  ((1+𝑥)/2)^(1/2)] Step-by-Step Explanation

Delve into the intricacies of inverse trigonometric functions with our detailed, step-by-step breakdown. In this video, we’ll guide you through solving complex problems involving [𝑐𝑜𝑠^(−1)  ((1+𝑥)/2)^(1/2)]  and differentiation, ensuring you grasp each concept thoroughly. Perfect for students and math enthusiasts looking to enhance their understanding.
#MathTutorial #InverseTrigFunctions #Calculus #Trigonometry #EducationalVideo


let's solve this complex derivative

problem step by step using the chain

rule we need to find the derivative of e

to the power of sin of sare < TK of X

let's break down the structure of this

function we have three nested functions

first the outer exponential function

then the S function in the middle and

finally the square root of x as the

innermost function to solve this we'll

use the chain rule multip times the

chain rule states that when

differentiating a composite function we

multiply the derivative of the outer

function by the derivative of the inner

function let's start with the first

application of the chain rule the

derivative of e to the power of anything

is e to that same Power Times the

derivative of what's inside now we need

to find the derivative of s of square <

TK X using the chain rule again

the derivative of s is cosine times the

derivative of what's inside finally

let's find the derivative of square < TK

of X the derivative of square < TK X is

1 / 2 < TK X now we can write our final

answer in a more organized way the

derivative equals cosine of s < TK of x

* e to the power of s of < TK X all / 2

< TK of x X let's verify that we didn't

miss anything we use the chain rule

three times first for the exponential

function then for the S function and

finally for the square root each

application gave us a term in our final

answer and that's how we find the

derivative of this complex function

using the chain rule multiple times

remember when dealing with nested

functions apply the chain rule for the

outside in and multiply all the terms

together

9.3-2.4 Mastering the Chain Rule Step-by-Step Differentiation Guide for (𝑒^𝑠𝑖𝑛√𝑥)

“In this video, we’ll dive deep into differentiating the composite function (𝑒^𝑠𝑖𝑛√𝑥) using the chain rule. Join us as we break down each step in detail, making it easy to follow and understand. Perfect for students looking to strengthen their calculus skills!”
Hashtags:
#Calculus #ChainRule #Differentiation #MathTutorial #MathHelp #hsc #hschighermath #1stpaper
#hschighermath1stpaper
#englishversion #cpmrana


let's solve this derivative step by step

using the product rule we want to find

the derivative of x^2 * cine X to solve

this we'll use the product rule the

product rule states that the derivative

of a product of two functions equals the

first function time the derivative of

the second plus the second function time

the derivative of the first let's

identify our functions let U of X = x^2

and V ofx = cosine X now let's find the

derivatives of each function separately

the derivative of x^2 is 2X and the

derivative of cosine X is sinx let's

substitute these into the product rule

we get x^2 * sinx plus cosine x * 2x now

let's simplify this expression first

we'll rearrange the terms to get 2X cine

x - x^2 sin x finally we can factor out

an X this gives us x * the quantity 2

cosine x - x sin x let's visualize how

this derivative works with a graph

here's the original function x^2 cine X

in blue and it's derivative in red to

summarize we use the product rule to

find the derivative of x^2 cosine X we

identified our functions found their

individual derivatives applied the

product rule and simplified our result

to get our final answer

Finding the Derivative of x^2 cosx  Using the Product Rule | Step-by-Step Explanation

In this video, discover how to differentiate x^2 cosx using the Product Rule in calculus.

What You'll Learn:
- Definition of the Product Rule
- Step-by-step differentiation of x^2 cosx
- Formula breakdown and example problem-solving

This example is ideal for calculus beginners and anyone reviewing the Product Rule!
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#Calculus #ProductRule #Differentiation #Derivative #MathTutorial #LearnMath #MathEducation #MathHelp