welcome to this lesson on logarithmic

differentiation today we'll be finding

the derivative of a complex expression

the quantity x to the X power all Rays

to the X power this is a challenging

problem because we have nested

exponentials let's solve it step by step

using logarithmic

differentiation step one first let's

simplify our expression using the

properties of exponents

when we have a power raised to another

power we multiply the exponents so x x

all raised to the X power simplifies to

X raised to x^2 step two let's call our

function y so y = x x^2 now we'll take

the natural logarithm of both sides this

is the key step in logarithmic

differentiation step three you using the

properties of logarithms we can bring

down the power the natural log of x raed

to a power equals that power times the

natural log of x step four now we'll use

implicit differentiation we'll take the

derivative of both sides with respect to

X on the left side we get 1 / y *

dydx on the right side we need to use

the product rule because we have x^2 *

lnx

step five let's simplify the right side

2x Ln X+ X which is x * the quantity 2

Ln x + 1 Step six now we can solve for

dydx we multiply both sides by y

remember that Y is our original

expression x to the x^2 and there we

have it the derivative of x to the X

power all rais to the X power is X x^2 *

x * 2 Ln x + 1 Let's highlight the key

steps we used first we simplified the

nested exponential then we used

logarithmic differentiation to make the

problem manageable and finally we used

implicit differentiation and the product

rule to find our answer this problem

demonstrates the power of logarithmic

differentiation in handling complex

exponential Expressions thank you for

watching

How to Find d/d[(𝒙^𝒙 )^𝒙] | Logarithmic Differentiation Made Easy

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

🔥 Master the art of logarithmic differentiation with this step-by-step tutorial on solving complex nested exponential derivatives.
In this video, we break down the challenging problem of finding the derivative of (xx)x using logarithmic differentiation. Perfect for calculus students and anyone preparing for advanced mathematics exams!
📚 What you’ll learn:
• Simplifying nested exponentials
• Applying logarithmic differentiation
• Using chain rule and product rule effectively
• Working with natural logarithm properties
• Step-by-step solution process
✏ Final Answer: d/dx(xx)x = x^(x²)[x(1 + 2ln(x))]
🎯 Level: Advanced Calculus / College Mathematics
📝 Related topics:
• Derivatives
• Exponential functions
• Natural logarithms
• Chain rule
• Product rule
#Calculus #Mathematics #Derivatives #MathTutorial #LogarithmicDifferentiation #ChainRule #ProductRule #MathHelp #MathEducation #CalculusTutorial


welcome today we'll solve a calculus

problem involving the differentiation of

a complex fraction our question is to

find the derivative of Ln of cine x / x

with respect to X to solve this we'll

need to use the quotient rule the

quotient rule states that for a fraction

U over V the derivative is V * d dxus u

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

and v in our case U = Ln of cosine X and

v = x now let's find D DX to

differentiate Ln of cosine X we'll need

to use the chain rule the derivative of

Ln of cine X is 1 / cosine x * the

derivative of cosine X which is sinx

finding dvdx is much simpler the

derivative of x is just one now now

let's substitute everything into the

quotient rule we have x * - Tan x - Ln

of cosine x * 1 all over x^2 and finally

we can simplify this expression our

final answer ISX Tan x - Ln of cine X

all over x^2 let's verify this is

correct by checking our steps we use the

quotient rule correctly found the

derivatives of both the numerator and

denominator accurately and simplified

our expression properly this completes

our solution

Mastering Derivatives: Differentiating [(ln(cos𝑥))/𝑥]  Step-by-Step

In this video, we dive deep into the process of differentiating the function  [(ln(cos𝑥))/𝑥] . Follow along as we break down each step using the quotient rule and chain rule. Perfect for students and anyone looking to strengthen their calculus skills. Don’t forget to like, subscribe, and hit the notification bell for more math tutorials!
#Calculus #Derivatives #MathTutorial #QuotientRule #ChainRule #LearningMath


welcome to this lesson on logarithmic

differentiation today we're going to

find the derivative of a particularly

challenging function y = e power of x to

X this is a complex exponential function

that would be very difficult to

differentiate using regular methods why

is this function so challenging well if

we try to use regular differentiation we

have to deal with both the chain Rule

and the product rule simultaneously

making it extremely complicated instead

we can use logarithmic differentiation

which will make this process much more

manageable let's solve the step by step

step one take the natural logarithm of

both sides this gives us Ln of Y = Ln of

e to power of x to X step two use the

logarith property that states Ln of e to

the power of something equals that

something this simplifies our equation

to Ln of y = x to

X step three take the natural log again

this gives us Ln of Ln of Y = Ln of x to

the x

step four use the logarithm property

that states Ln of a to the power of b

equal B * Ln of a this gives us Ln of Ln

of y = x * Ln of

X step five now we can differentiate

both sides with respect to X on the left

side we'll use the chain rule the

derivative of Ln of Ln of Y with respect

to x = 1 Ln of y * 1 / y *

dydx step six on the right side we use

the product rule the derivative of x Ln

X = Ln x + x * 1 /x which simplifies to

Ln x + 1

step seven now we can solve for

dydx multiply both sides by Y and Ln of

Y this gives us dydx = y * Ln of y * the

quantity Ln x +

1 step8 finally substitute back yal e to

power of x x this gives us our final

answer the derivative equals e to power

of x to x * x x * the quantity Ln x +

1 let's review what we did we used

logarithmic differentiation to handle

this complex function by taking the

natural log twice applying logarithm

properties differentiating both sides

using the chain Rule and product rule

and then substituting back our original

function this technique is powerful for

handling complex exponential and product

functions thank you for watching

How to Find d/dx[〖𝒆^𝒙〗^𝒙] | Logarithmic Differentiation Made Easy

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

Master the art of logarithmic differentiation with this comprehensive tutorial on finding the derivative of 〖𝒆^𝒙〗^𝒙. Perfect for calculus students and math enthusiasts!
In this video, we’ll cover:
📚 Why regular differentiation won’t work here
🔍 The power of logarithmic differentiation
✍ Step-by-step solution process
💡 Important logarithm properties
🎯 Chain rule application
✅ Final answer verification
🔔 Subscribe for more calculus tutorials!
❓ Leave your questions in the comments below
#Calculus #Mathematics #Derivatives #LogarithmicDifferentiation #MathTutorial #CalculusTutorial #MathEducation #APCalculus #MathHelp


let's solve this differentiation problem

step by step we need to find the

derivative of y = sin 2x over 1 + cosine

2X all squared let's break this down

into manageable steps first let's recall

some important trigonometric identities

that we'll need let's start with the

numerator using the double angle formula

for S we can replace sin 2x with 2 2 sin

x cine X next let's look at the

denominator using another double angle

formula 1 + cosine 2X = 2 cosine 2 x now

let's expand this expression in the

numerator we get 4 sin^2 x cine 2 x in

the denominator we get 4 cosine to the

4th X the fours cancel out in the

numerator and denominator

this leaves us with sin^2 X over cosine

2 x this is a key moment we can

recognize that sin^2 X over cosine 2 x

is equal to tangent 2 x now we're ready

to differentiate we need to find d y DX

= d/ DX of tangent 2 x using the chain

rule we multiply by 2 tangent X and then

multiply by the derivative of tangent X

which is secant 2 x and this is our

final answer the derivative of sin 2x

over 1 + cosine 2X all 2ar is 2 tangent

X secant 2 x let's review the key steps

we use double angle formulas simplified

the fraction recognize the tangent squar

pattern and then use the chain rule to

differentiate understanding these

patterns and identities is crucial for

solving trigonometric differentiation

problems

Mastering Trigonometric Differentiation [(sin2𝑥/(1+cos2𝑥 ))^2] Step-by-Step Guide

Unlock the secrets of trigonometric differentiation with our detailed, step-by-step tutorial. Learn how to differentiate complex trigonometric functions like y=[(sin2𝑥/(1+cos2𝑥 ))^2]  with ease. This video breaks down each step, providing clear explanations along the way. Perfect for students and anyone looking to strengthen their calculus skills. Subscribe for more insightful math tutorials!
#Calculus #Differentiation #Trigonometry #MathTutorial #LearnMath #Mathematics


let's solve this complex differentiation

problem step by step we start with the

expression cosine of natural log X Plus

natural log of tangent X to

differentiate this we'll use the chain

Rule and split it into two parts the

derivative of the first part cosine of

natural log X plus the derivative of the

second part natural log of tangent X

let's handle each part separately for

the first part

using the chain rule we get Negative s

of natural log X time the derivative of

natural log X which is 1 /x for the

second part we use the chain rule again

the derivative of natural log of tangent

x = 1 / tangent x * the derivative of

tangent X which is secant 2 x now we

combine these parts we have negative s

of natural log x/ X + 1 / Tang x * sec^2

X let's simplify the second term

remember that tangent X = sin x over

cosine X and secant 2 x = 1 cosine 2 x

substituting these in we get Negative s

of natural log x/ X + cosine x/ sin x *

1 cosine 2 x this simplifies to negative

s of natural log x/ x + 1 / sin x * 1 /

cosine X the second term can be

Rewritten as 2/ 2 sinx cosine X which

equal 2 over sin of 2x using the double

angle formula finally we can write this

as 2 cose 2x - 1 /x * sin of natural log

X which is our final answer this problem

demonstrated the use of chain rule

product Rule and trigonometric

identities and calculus remember to

break down complex problems into simpler

steps and use your knowledge of

trigonometric relationships

Differentiating Tricky Composite Functions {cos(𝑙𝑛𝑥)+ln(𝑡𝑎𝑛𝑥)} Step by Step Guide

Learn how to differentiate the function {cos(𝑙𝑛𝑥)+ln(𝑡𝑎𝑛𝑥)}  with this detailed, step-by-step guide. We’ll cover the application of the chain rule, sum rule, and simplification using trigonometric identities to help you master these concepts. Perfect for students and math enthusiasts aiming to strengthen their calculus skills!
#Calculus #Differentiation #MathTutorial #Derivatives #ChainRule


welcome today we'll solve a challenging

derivative problem involving

trigonometric functions and substitution

let's start by looking at our problem

we're given that y = the quantity 1 -

x^2 over 1 + x^2 all squared and we're

also given that x equal tangent Theta

our goal is to find

dydx this is a complex expression so

let's break it down into steps using the

chain rule we can write

dydx as Dy D Theta / DX D Theta let's

find DX D Theta first since it's simpler

since x equal tangent Theta DX D Theta

equal secant 2 Theta now for Dy D Theta

we need to recognize that the expression

1 - x^2 over 1 + x^2 is actually equal

to cosine 2 th and when we Square it we

get c^ 2 2 th when we differentiate

cosine 2 2 Theta with respect to Theta

we get -4 cosine 2 Theta sin 2 Theta now

we can substitute these into our chain

rule equation we get -4 cosine 2 thet

sin 2 thet /^ 2 thet to simplify this

further we'll use these identities

cosine 2 thet = 1 - t^ 2 thet over 1 +

Tang 2 thet and sin 2 thet = 2 Tang thet

over 1 + Tang 2 thet substituting these

in and simplifying we get -4 * 1 - Tang

2 th over 1 + t^ 2 thet * 2 Tang thet

over 1 + Tang

* 1 1 + Tang 2 this simplifies to 8 Tang

thet * 1 - Tang 2 Theta all over 1 +

Tang 2 Theta cubed finally since X =

tangent Theta we can write our answer as

-8x * 1 - x^2 all over 1 + x^2 cubed and

there we have it we've successfully

found dydx X using trigonometric

substitution and the chain rule

Understanding Calculus: Derivatives Using Trigonometric Substitutions [((1−𝑥^2)/(1+𝑥^2 ))^2 ]

Question
Given  y= [((1−𝑥^2)/(1+𝑥^2 ))^2 ] and , find dy/dx .

Explore the step-by-step process of finding derivatives using trigonometric substitutions. In this tutorial, we’ll go through the detailed calculation of  dy/dx . for a given function, leveraging identities and calculus techniques. Perfect for students looking to understand complex derivatives in calculus!
#Calculus #Derivatives #Trigonometry #MathTutoring #STEM


welcome today we'll solve a challenging

derivative problem involving inverse

trigonometric functions let's start with

our problem we need to find the

derivative of y = inverse tangent of X

over theun of 1 - x^2 this looks

complicated but don't worry we can solve

it step by step using a clever

substitution let's make the substitution

x equals s of T this means tal inverse s

of X now let's substitute this into our

original expression we replace x with s

of T in the numerator and in the

denominator we get the square root of 1

- sin^2 t here's where we can use a

fundamental trigonometric identity

remember that 1 - sin^2 t = cine 2 T

using this ident

we can simplify our expression to

inverse tangent of sin t/ cosine T now

here's the key Insight sin T over cosine

T is the definition of tangent T so our

expression becomes inverse tangent of

tangent T when we have inverse tangent

of tangent T these functions cancel out

because they're inverse functions so our

expression simplify to just

T remember though t equal inverse sin of

X so our expression is really y =

inverse sin of X now we can find the

derivative the derivative of inverse sin

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

therefore the derivative of inverse

tangent of x/ < TK of 1 - x^2 = 1 / < TK

of 1 - x^2 this problem shows how making

a clever substitution can help simplify

a complex derivative the key was

recognizing that we could substitute xal

sin of T which L us use trigonometric

identities to simplify the expression

before taking the derivative

Mastering Derivatives: Unraveling Inverse Trigonometric Functions [𝑡𝑎𝑛^(−1)   𝑥/√(1−𝑥^2 )]

Question
Find the derivative of the function [𝑡𝑎𝑛^(−1)   𝑥/√(1−𝑥^2 )] .
Dive into the world of calculus with this tutorial on differentiating inverse trigonometric functions. Follow along as we simplify and differentiate the expression  [𝑡𝑎𝑛^(−1)   𝑥/√(1−𝑥^2 )] step-by-step. Ideal for students and anyone looking to strengthen their understanding of calculus concepts!
#Calculus #Derivatives #Trigonometry #MathTutorial #InverseFunctions


let's solve this differentiation problem

step by step using the quotient rule we

need to differentiate the function f of

T = 1 + t + T ^2 / 1 - t + T ^2 let's

recall the quotient rule the derivative

of U over V equal V * DDT minus U *

dvdt all over V ^2 first let's identify

u and v here U equal 1 + t + T ^2 and V

= 1 - t + T ^2 next we need to find do

DT and dvdt the derivative of U is 1 +

2T and the derivative of V is - 1 + 2T

now let's substitute everything into the

quotient rule formula we get 1 - t + T

^2 * 1 + 2T - 1 + t + t^ 2 * - 1 + 2T

all over 1 - t + t^ 2^ 2 let's expand

the numerator first let's multiply 1 - t

+ T ^ 2 by 1 + 2T this gives us 1 + 2 t

- t - 2 T ^2 + T ^ 2 + 2 T Cub which

simplifies to 1 + t + 2 t^ 2 - 2 T Cub

now for the second part let's multiply 1

+ t + T ^2 by -1 + 2T this gives us -1 +

t - 3T ^2 + 2 T cubed now we can

subtract these Expressions 1 + t + 2 T

^2 - 2 T Cub - -1 + t - 3 T ^2 + 2 T Cub

this simplifies to 2 - 2 T ^2 therefore

our final answer is 2 *

1 - t ^ 2/ 1 - t + T ^ 2^ 2 and that's

how we use the quotient rule to

differentiate this rational function

Quotient Rule Masterclass: Differentiating Complex Fractions [(𝟏+𝒕+𝒕^𝟐)/(𝟏−𝒕+𝒕^𝟐)] 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+𝒕+𝒕^𝟐)/(𝟏−𝒕+𝒕^𝟐)] , 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 complex trigonometric

differentiation problem step by step our

question is to find the derivative of

the square < TK of 1 + cosine X over 1 -

cine x with respect to X let's break

this down into manageable steps first

we'll use the half angle identities to

rewrite the numerator and denominator

remember 1 + cosine x = 2 cosine 2 x / 2

and 1 - cosine x = 2 sin x / 2 next we

can simplify this fraction under the

square root the twos cancel out and what

we're left with is cosine 2 x/ 2 / sin 2

x/ 2 this is the definition of cotangent

2 x/ 2 the square root of Cent squar is

just coent itself so our expression

simplifies to the derivative of coent x

/ 2 now we can use the chain rule

remember that the derivative of coent is

netive cosecant SAR so we'll have

negative cose 2 of X over 2 * the

derivative of x over 2 the derivative of

x over 2 is simply 12 so we can multiply

that through cose squ is the same as 1 /

sin 2 so we can rewrite our our

expression as -1/2 sin^2 x / 2 finally

we can use another half angle identity

sin^2 x/ 2 = 1 - cosine x / 2 when we

substitute this in and simplify we get

our Final Answer -1/ 1 - cine X this

problem demonstrates the importance of

knowing your trigonometric identities

and being comfortable with the chain

rule by breaking down the problem into

smaller steps and using these tools we

were able to find the derivative of this

complex expression

Mastering Trigonometric Derivatives (√((1+cos𝑥)/(1−cos𝑥 ))) A Step-by-Step Guide

Dive into the fascinating world of calculus with our comprehensive video on differentiating complex trigonometric functions. In this tutorial, we’ll guide you through a detailed step-by-step process to differentiate the function (√((1+cos𝑥)/(1−cos𝑥 ))). By breaking down each step and explaining the trigonometric identities involved, you’ll gain a clear understanding of the concepts needed to tackle similar problems. Don’t miss out on mastering this essential calculus skill!
#Calculus #Trigonometry #MathTutorial #Differentiation #MathHelp


let's solve this challenging derivative

problem step by step we need to find the

derivative of y equal inverse tangent of

1 - < TK X over 1 + < TK X the key to

solving this problem is to recognize

that we can rewrite this as a difference

of inverse tangents there's a useful

identity that states inverse tangent of

a minus inverse tangent of of b equal

inverse tangent of a minus b/ 1 + a

looking at our fraction we can rewrite

it as inverse tangent of 1 minus inverse

tangent of squ < TK X this is because

when we substitute a equals 1 and be

equal sare < TK of X into the identity

we get our original expression now we

know that tangent of pi over 4 equal 1

therefore inverse tangent of 1 = < / 4

this allows us to simplify our

expression to Y = < / 4 - inverse

tangent of s < TK X now we can

differentiate this expression the

derivative of pi over 4 is zero since

it's a constant for the inverse tangent

term we'll use the chain Rule and the

fact that the derivative of inverse

tangent of U is 1 1 + u^2 * DDX the

derivative of square < TK X is 1 / 2 <

TK X substituting this in we get -1 1 +

x * 1 2 otk of X finally we can combine

these fractions to get our final answer

-1/2 square root of x * 1 + x to verify

this is correct we can check that this

matches our original problem the

derivative of inverse tangent of 1 - <

TK x/ 1 + < TK x

= 1 / 2 < TK of x * 1 + x

Mastering Derivatives: Inverse Tangent Function Simplified! [𝑦=𝑡𝑎𝑛^(−1)   (1−√𝑥)/(1+√𝑥)]

Question
Find the derivative of the function [𝑦=𝑡𝑎𝑛^(−1)   (1−√𝑥)/(1+√𝑥)].
Unlock the secrets to differentiating inverse trigonometric functions with this step-by-step tutorial. We’ll walk you through the process of finding the derivative of  [𝑦=𝑡𝑎𝑛^(−1)   (1−√𝑥)/(1+√𝑥)] . Perfect for calculus students looking to strengthen their understanding. Don’t miss out—subscribe for more math insights!
#Calculus #MathTutorial #Derivatives #InverseTrigonometry #STEM


welcome today we'll work through a

complex derivative problem step by step

let's start by looking at our expression

here's our initial expression that we

need to differentiate we need to find

the derivative of x over 1 + the < TK of

1 - x^2 all raised to the nth power

let's break this down step by step first

we'll use the chain rule since we have a

function to a power this gives us n *

our function to the N -1 power time the

derivative of the inner function now for

the inner derivative we'll need to use

the quotient rule remember the quotient

rule states that the derivative of U

over V equal V * the derivative of uus U

* the derivative of V all over V ^2

applying this to our problem we get this

expression the the numerator has 1 + <

TK of 1 - x^2 * the derivative of x - x

* the derivative of our denominator

let's evaluate these derivatives the

derivative of x is simply one and for

the denominator we need to use the chain

rule to find the derivative of square <

TK of 1 - x s after simplifying the

numerator we get this expression no

notice how the terms combine nicely in

the numerator through further algebraic

manipulation we can combine terms and

simplify our expression to this form and

finally we arrive at our final answer

the derivative equals n /x * the < TK of

1 - x^2 time our original expression

raise to the nth power this problem

demonstrates the power of combining

different differentiation rules the

chain Rule and quotient rule along with

careful algebraic manipulation to arrive

at an elegant final result

Mastering Derivatives [𝑥/(1+√(1−𝑥^2 ))]^𝑛  Step-by-Step Guide to Differentiation

Question
Given the expression: [𝑥/(1+√(1−𝑥^2 ))]^𝑛  calculate its derivative with respect to  x.

Unlock the secrets of calculus with our detailed walkthrough on differentiating complex expressions. From using the chain rule to applying the quotient rule, this video provides a clear, step-by-step explanation to help you master derivatives. Perfect for students and calculus enthusiasts.

#Calculus #Mathematics #Derivatives #Education #Differentiation


let's find the derivative of sin x *

cine X using the product rule first

let's write out what we're trying to

find for this problem 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 equal sinx nval cosine X

let's substitute these into the product

rule formula

we get sin x * the derivative of cosine

X plus cosine x * thetive of sin x now

let's recall the derivatives of s and

cosine the derivative of cosine X is

sinx and the derivative of sinx is cine

X let's substitute these derivatives

into our equation we get sinx * sinx

plus cosine x * cosine X now now let's

simplify sin x * sin x is sin^2 X and

cosine x * cosine X is cosine 2 x we can

rearrange this as cosine 2 x - sin^2 x

finally we can use a double angle

identity the expression c^ 2 x - sin^2 x

is equal to cosine of 2x therefore the

derivative of sin x * cosine x = cosine

of 2x to verify this visually let's look

at the graphs here's sin x * cosine X in

blue and its derivative cosine of 2x in

red and that's how we find the

derivative of sin x * cine X using the

product rule and simplify it to cine of

Finding the Derivative of (sin𝑥 cos𝑥) Using the Product Rule | Step-by-Step Explanation

In this video, learn to differentiate (sin𝑥 cos𝑥) using the Product Rule in calculus.

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

This example is perfect for calculus beginners or those reviewing the Product Rule!

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


let's work through finding the

derivative of e to the power of sin 2x

step by step first let's look at our

original function we have eras to the

power of sin 2x this is a composite

function we have an exponential function

on the outside and a trigonometric

function on the inside to differentiate

this we'll need to use the chain rule

the chain rule states that when we have

a composite function f G of X the

derivative is equal to frime of G ofx *

G Prime of X let's start our

differentiation first we write out that

we're taking the derivative with respect

to X next using the chain rule we know

that when we differentiate e to a power

we keep the E to that power and multiply

by the derivative of what's inside now

we need to differentiate sin 2x remember

the derivative of s is cosine but we

also need to account for the chain rule

again because of the 2x inside the

derivative of 2x is simply 2 finally we

can rearrange this to write our answer

as 2 * cosine 2X * e^ of sin 2x let's

visualize what happened here we started

with an exponential function with a

trigonometric function inside using the

chain rule we one kept the original

exponential function to multiply by the

derivative of the inner sign function

which gave us cosine 3 and multiply by

the derivative of the argument of s

which was to this gave us our final

answer and that's how we find the

derivative of e to the power of sin 2x

using the chain rule thanks for watching

Mastering the Chain Rule Differentiating [(𝑒^sin2𝑥)]  Step by Step!

Question: Find the derivative of  [(𝑒^sin2𝑥)]
In this video, we break down the process of differentiating the function  (𝑒^sin2𝑥)  using the chain rule. Whether you’re a calculus student or just looking to brush up on your differentiation skills, this tutorial will guide you through each step with clear explanations and examples. Master the art of using the chain rule and enhance your math skills today!
#Calculus #ChainRule #Differentiation #MathTutorial #ExponentialFunctions #MathHelp #hsc #hschighermath #1stpaper
#hschighermath1stpaper
#englishversion


let's solve a calculus problem involving

the differentiation of trigonometric

functions with degree measures our

question is to find the derivative of

thet degre time sin of thet degrees with

respect to Theta before we start

differentiating we need to convert the

degrees to radians we do this by

multiplying Theta degrees by pi over

180 this gives us Theta pi over 180

now we can apply the product rule

remember the product rule states that

the derivative of U * V equal U * the

derivative of V + V * the derivative of

U in our case U is Theta pi over

180 and V is s of theta pi over

180 let's handle each part separately

for the first term we need to use the

chain rule when differentiating sign the

derivative of s is cosine and we

multiply by the derivative of the inner

function this gives us Theta pi over 180

* cosine of theta pi over 180 * piun

over 180 for the second term we just

need to differentiate Theta pi over 180

with respect to Theta which gives us pi

over

180 now we can combine like terms let's

factor out pi over

180 from both terms this gives us our

final answer > over 180 * the quantity

Pi Theta over 180 * cine of theta pi/

180 + sin of theta pi over

180 let's verify this makes sense our

answer contains both s and cosine terms

which we would expect from

differentiating a sign function the pi

over 180 terms appear because we

converted from degrees to radians and

the Theta term remains because we

started witha * sin thet and that's how

we differentiate a trigonometric

function involving degree measures

remember to always convert to radians

first then apply the product rule and

chain rule as needed

Mastering Calculus: Differentiating Trigonometric Functions (𝜽° 𝒔𝒊𝒏𝜽°)

Question: Differentiate the expression  (𝜽° 𝒔𝒊𝒏𝜽°)  with respect to 𝜽.
Unlock the secrets to differentiating trigonometric functions with our step-by-step guide! In this video, we break down the differentiation process for expressions involving trigonometric components, using the product and chain rules. Perfect for students or anyone wanting to refine their calculus skills.
#Calculus #MathTutorial #Trigonometry #Differentiation #STEMLearning


let's solve a challenging derivative

problem involving inverse trigonometric

functions our goal is to find the

derivative of tan inverse of 2 < tkx

over 1 - x this looks complicated but we

can solve it using a clever substitution

let's substitute Square < TK of X = Tan

t therefore T = tan inverse of s < tkx

using this substitution

our expression becomes tan inverse of 2

tan t/ 1 - tan^ 2 T this can be

simplified to tan inverse of tan of 2T

using the double angle formula for

tangent now since tan inverse and tan

are inverse functions they cancel out

leaving us with Just 2 T remember T is

tan inverse of < TK X so we can rewrite

this as 2 * tan inverse of < TK X to to

differentiate this we'll use the chain

rule the derivative of tan inverse U is

1/ 1 + u^2 * the derivative of U here U

is s < TK X so we get 2 over 1 + x * 1 2

< tkx finally we can simplify this to 1

/ < tkx * 1 + x to verify this is

correct we could check that this

derivative matches what we would get

using implicit differentiation but

that's beyond the scope of this video

let's review the key steps first we made

a clever substitution to simplify the

expression then we used the double angle

formula for tangent next we simplified

using the inverse property of tangent

functions finally we use the chain rule

to find the derivative and simplified

our answer

Exploring Derivatives: Inverse Tangent Function [𝑡𝑎𝑛^(−1)   (2√𝑥  )/(1−𝑥)] Explained Step-by-Step

Question: Find the derivative of [𝑡𝑎𝑛^(−1)   (2√𝑥  )/(1−𝑥)] .
Dive into the world of calculus with a detailed explanation of how to find the derivative of an inverse tangent function. Using clear, step-by-step methods, this video breaks down each part of the process to enhance your understanding and problem-solving skills. Perfect for students and enthusiasts looking to master calculus concepts.
#Calculus #Derivatives #InverseTrigonometry #MathTutorial #Education #StudyMath


welcome today we'll solve a challenging

derivative problem using logarithmic

differentiation our goal is to find the

derivative of this function y = x Cub *

the < TK x^2 + 4 over x^2 + 3 this

function looks quite complicated to

differentiate directly so instead let's

take the natural logarithm of both sides

this will allow us to use the properties

of logarithms to break down this complex

expression now we can use the properties

of logarithms to break this down

remember the logarithm of a product is

the sum of the logarithms and the

logarithm of a quotient is the

difference of logarithms let's apply

these properties first we separate the X

Cub term then the square root becomes a

1/2 power of the fraction next we can

further simplify this using the power

property of logarithms the X Cub becomes

3 *

lnx and the square root becomes 1/ 12 *

the logarithm of the fraction now we can

split the fraction inside the logarithm

giving us three terms now comes the key

step we take the derivative of 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 to solve for dydx we

multiply both sides by y finally we

substitute back the original expression

for y to get our final answer this gives

us the derivative of our original

function and there you have it we've

found the derivative using logarithmic

differentiation this method is

particularly useful when dealing with

products and quotients of functions as

it turns multiplication and division

into addition and subtraction

Calculating Derivatives Using Logarithmic Differentiation [𝑥^3 √((𝑥^2+4)/(𝑥^2+3))]

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

Learn how to find the derivative of complex functions using logarithmic differentiation. In this tutorial, we break down each step with clear explanations. Ideal for students wanting to master calculus concepts.
#Calculus #LogarithmicDifferentiation #Derivatives #MathTutorial #LearnMath


let's find the derivative of the natural

logarithm of the square < TK of X we

start with Ln of square < TK of x to

solve this we'll need to use the chain

rule the chain rule tells us that when

we have a composite function like this

one we need to multiply the derivative

of the outer function by the derivative

of the inner function in our case the

outer function is the natural logarithm

and the inner function is the square

root of x remember the derivative of Ln

X is 1 /x let's apply the chain rule

step by step first we write one over the

inner function which is square < TK of X

then multiply by the derivative of the

inner function now we need to find the

derivative of square otk of X remember

that square < TK of X is the same as x

to the 1/2 power the derivative of x to

the 1/2 is2 * x^ -2 power which

simplifies to 1 / 2 < TK X let's

substitute this back into our chain rule

equation we get 1 < TK x * 1 / 2 < TK X

when we multiply these fractions we get

1/ 2X and that's our final answer the

derivative of Ln of squ < TK X is 1 over

2x let's review what we did we use the

chain rule to break down the derivative

then found the derivative of the inner

function using the power rule and

finally combined our fractions to get

our answer

Understanding the Derivative of Logarithmic Functions [(𝑙𝑛√𝑥)] Step-by-Step Guide

Question: Find the derivative of [(𝑙𝑛√𝑥) ]
Dive into the fascinating world of calculus as we explore how to find the derivative of the logarithmic function (𝑙𝑛√𝑥). Learn about the chain rule and power rule in a simple and intuitive way. Perfect for students and enthusiasts wanting to grasp the fundamentals of differentiation. Don’t forget to like, subscribe, and comment below with your thoughts!
#englishversion #cpmrana #Calculus #Differentiation #LogarithmicFunctions #MathTutorial #ChainRule


let's solve this challenging derivative

problem step by step our goal is to find

the derivative of inverse tangent of

secant X+ tangent X let's start by

applying the chain rule remember the

derivative of inverse tangent of U is 1

1 + U ^2 times the derivative of U in

our case U is secant x + tangent X now

we need to find the derivative of U

which is the derivative of secant x +

tangent X the derivative of secant x is

secant x tangent X and the derivative of

tangent X is secant 2 x adding these

together gives us secant x tangent X

Plus secant 2 x let's substitute

everything back into our original

formula we get 1 over 1 + the quantity

secant x + tangent x^2 * secant x

tangent X x +^ 2 x now let's expand the

denominator when we Square secant x +

tangent X we get secant 2 x + 2 secant x

tangent x + tangent 2 x remember that

tangent 2 x = secant 2 X - 1 using this

identity we can simplify our denominator

the denominator becomes 2 secant s x + 2

secant x tangent X looking at our

fraction now we can factor out secant 2

x + secant x tangent X from both

numerator and denominator and finally

after canceling the common factors we

arrive at our answer 1/2 this problem

demonstrated the importance of knowing

your derivative formulas trigonometric

identities and algebraic manipulation

skills

Mastering Derivatives: Simplifying {𝑡𝑎𝑛^(−1) (sec𝑥+tan𝑥 )} Step-by-Step

Question
Simplify the derivative of the function {𝑡𝑎𝑛^(−1) (sec𝑥+tan𝑥 )}.

Unlock the mystery behind simplifying derivatives with this comprehensive walkthrough of  {𝑡𝑎𝑛^(−1) (sec𝑥+tan𝑥 )}. Perfect for calculus students and math enthusiasts, this video breaks down the process into easy-to-understand steps. Follow along for clear explanations and valuable insights to help you ace your calculus course.
#Calculus #Derivatives #MathHelp #LearningMath #Trigonometry #MathTutorial


let's solve this calculus problem step

by step we need to find the derivative

of e to x * tangent X first let's

identify that this is a product of two

functions we have U of xal e to X and V

ofx = tangent X since this is a product

of two functions we'll 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 let's

apply this to our problem we get e to

the x * the derivative of tangent X Plus

tangent x * the derivative of e to X now

let's find each derivative separately

first the derivative of tangent X is

secant squ x and the derivative of e to

the x is just e to the x substituting

these derivatives back into our equation

we get e to x * secant 2 X Plus tangent

x * 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 secant 2 x + tangent X let's

review the key steps we identified this

as a product rule problem apply app the

product rule formula found the

derivatives of tangent X and E to X

separately substituted these back in and

finally factored out the common term e

to X and that's our solution

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

In this video, learn to differentiate 𝑒^𝑥  tan𝑥 using the Product Rule in calculus.

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

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


let's solve this calculus problem step

by step using the quotient rule we need

to find the derivative of x sin x / 1 +

cosine X let's recall the quotient rule

formula the derivative of U over v = v *

the derivative of U minus U * the

derivative of V all / V ^2 let's

identify our u and v in this case U

equal x sinx and V = 1 + cine X now we

need to find the derivatives of u and v

for U which is X sinx will use the

product rule this gives us sin x + x

cine x 4 V which is 1 + cinex the

derivative is sinx let's substitute

these into our quotient rule formula we

get 1 + cosine x * sin x + x cosin x - x

sin x * sin x all over 1 + cosine x^ 2

let's expand the numerator first

distribute 1 + cosine X we get sin x +

sin x cosine x + x cosine x + x cosine ^

2 x then for the negative

term * sinx gives us plus x sin^2 x now

let's use a trigonometric ident

we know that sin^2 x + cine 2 x = 1 so x

* sin^2 x + x * cosine 2 x = just X this

simplifies our expression to sin x + sin

x cine x + x cosine x + x we can Factor

this as sin x * 1 + cine x + x * 1 +

cine X finally we can cancel 1 + cosine

X from the numerator and denominator

giving us our final answer x + sinx over

1 + cine X and that's how we use the

quotient rule to find this derivative

remember to take it step by step and use

your trigonometric identities when they

help simplify the expression

Differentiating Complex Fractions: Simplifying Trigonometric Expressions [(𝒙𝒔𝒊𝒏 𝒙)/(𝟏+𝒄𝒐𝒔 𝒙)]

Explore the step-by-step process of differentiating and simplifying trigonometric expressions such as [(𝒙𝒔𝒊𝒏 𝒙)/(𝟏+𝒄𝒐𝒔 𝒙)]. Follow along to simplify it to [( (𝒙+𝒔𝒊𝒏 𝒙))/((𝟏+𝒄𝒐𝒔 𝒙) )] . Perfect for calculus students looking to enhance their skills in handling complex fractions and trigonometric derivatives.
#Trigonometry #Calculus #MathTutorial #Differentiation #LearnMath


let's solve this complex derivative

problem step by step we need to find the

derivative of tan inverse of the square

< TK of 1 - x over 1 + x this is a

challenging problem that we'll solve

using substitution let's make the

substitution X = cine of T which means T

= inverse cosine of x after making this

substitution we can rewrite our

expression the expression under the

square < TK becomes 1 - cine T over 1 +

cine T now comes a key inside we can use

the double angle formula which states

that 1 - cosine T = 2 sin^2 T / 2 and 1

+ cine T = 2 cosine 2 of T / 2 applying

these identities our expression becomes

tan inverse of the square < TK of 2

sin^2 t / 2 / 2 cosine ^ 2 t/ 2 the twos

cancel out and under the square root

we're left with sin^2 t / 2 / cosine 2

t/ 2 which is just tangent 2 of t/ 2 the

square root and the square cancel each

other giving us tan inverse of tangent T

/ 2 here's another key Insight when you

have tan inverse of tangent of an angle

they cancel out leaving you with just

the angle so our expression simplifies

to T / 2 remember that tal inverse

cosine of x so we can write y = 12

inverse cosine of x now we can

differentiate the derivative of inverse

cosine of x is -1 over the < TK of 1 -

x^2 therefore our final answer is -1 /2

* 1 the < TK of 1 - x^2 which simplifies

to 1 /2 * < TK of 1 - x squar this

problem showcases the power of

trigonometric substitution and

identities in solving complex

derivatives the key was recognizing that

we could simplify our expression using

the half angle formulas and the

properties of inverse trigonometric

functions

Mastering Derivatives: ArcTan and Trigonometric Simplifications [𝑡𝑎𝑛^(−1) √((1−𝑥)/(1+𝑥))]

Question
Find the derivative of the function [𝑡𝑎𝑛^(−1) √((1−𝑥)/(1+𝑥))] .

Unlock the secrets of calculus with our step-by-step guide to finding derivatives involving arc tangents and trigonometric expressions. This video breaks down the process of differentiating [𝑡𝑎𝑛^(−1) √((1−𝑥)/(1+𝑥))] with detailed explanations and simplifications. Perfect for students and math enthusiasts aiming to deepen their understanding of calculus concepts.
#Calculus #Derivatives #MathTutorial #Trigonometry #ArcTan #STEM


welcome today we're going to learn about

logarithmic differentiation by finding

the derivative of yal x to the power of

1 /x this is a particularly tricky

function because X appears in both the

base and the exponent let's try to

understand why this is challenging to

differentiate directly if we try to use

the power rule we'd run into trouble

because the exponent contains X if we

try to use the exponential rule we'd

have issues because the base also

contains X this is where logarithmic

differentiation comes in handy step one

let's take the natural logarithm of both

sides this gives us Ln of Y = Ln of X

to^ of 1 /x step two now we can use the

logarithm property that says Ln of a to

the power of Bal B * Ln of a this

simplifies our equation to Ln of Y = 1

/x * Ln of X step three now we can

differentiate both sides with respect to

X the left side becomes 1 / y * dydx

because we're using the chain rule for

the right side we need to use the

product rule since we have two functions

being multiplied together step four

let's hand handle the right side

carefully using the product rule

remember the product rule says that the

derivative of U * v = u * dvdx + V * DDX

in our case U is 1 /x and V is Ln of X

now let's evaluate each derivative the

derivative of 1 /x is NE 1 /x^2 and the

derivative of Ln of X is 1/x

substituting these in step five now we

can solve for

dydx multiply both sides by Y and

remember that y = x to^ of 1 /x this

gives us our final answer Dy DX = x^ of

1x * the 1 - Ln of X over x^2 let's

summarize why logarithmic

differentiation was helpful here

by taking the natural log first we were

able to bring down the exponent and

create a much simpler expression to

differentiate this technique is

particularly useful when dealing with

functions where variables appear in both

the base and exponent or with products

that would be difficult to differentiate

directly and that's how we use

logarithmic differentiation to find the

derivative of y = x to the power of 1 /x

thank you for watching

Master Logarithmic Differentiation: Solving [y = x^(1/x)] Step by Step

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

📚 In this comprehensive tutorial, we break down how to find the derivative of y = x^(1/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! ⭐
Hashtags:
#Calculus #Mathematics #LogarithmicDifferentiation #MathTutorial #CalcTutorial #MathEducation #LearnMath #Derivatives


let's solve this challenging

differentiation problem step by step we

need to find the derivative of 10 Rays

to the natural log of sinx this is a

complex function that combines

exponential logarithmic and

trigonometric functions let's break down

how we'll solve it first we recognize

this follows the pattern of

differentiating a function of the form a

to the U where is 10 and U is natural

log of sinx four functions of this form

we know that the derivative equals a to

the U time the natural log of a * the

derivative of U with respect to X

applying this rule to our problem we get

10^ the natural log of sinx time the

natural log of 10 time the derivative of

natural log of sinx with respect to X

now we need to find the derivative

of natural log of sinx using the chain

rule this equals 1 / sin x * the

derivative of sinx substituting in the

derivative of sin x which is cosine X we

get 10 to the natural log of sin x *

natural log of 10 * cine X over sin x

finally we can simplify this further by

recognizing that cine X over sinx is the

definition of coent X therefore our

final answer is 10 the natural log of

sinx * natural log of 10 time Cent X

let's review the key steps we identified

the pattern of a to the U applied the

chain rule for logarithmic

differentiation and simplified our

expression using trigonometric

identities to reach our final answer

Logarithmic Differentiation (10^(lnsin𝑥)) Step by Step Guide

Unlock the secrets of logarithmic differentiation! In this video, we break down the process of differentiating complex expressions such as (10^(lnsin𝑥))  into simple, understandable steps. Whether you’re a calculus student, teacher, or math enthusiast, you’ll find our detailed walkthrough helpful in mastering this essential mathematical technique. Join us and boost your differentiation skills today!
#Calculus #Differentiation #MathTutorial #LogarithmicDifferentiation #Trigonometry


let's solve this calculus problem step

by step using the quotient rule we need

to find the derivative of cosine x/ x^2

+ sinx first let's recall the quotient

rule formula 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 functions U equal

cine X and V = x^2 + sin x now let's

find the derivatives of u and v the

derivative of cine X is sinx and the

derivative of x^2 + sin x is 2x + cosine

X let's substitute these into the

quotient rule formula we get x^2 + sin x

* sinx - cinx * 2x + cosin x all over

x^2 + sin x^2 now let's expand the

numerator first let's multiply x^2 + sin

x by sin x this gives us x^2 sin x -

sin^2 x next we multiply cosine X by 2x

+ cosine X getting 2x cosine x + cosine

2 x let's factor out the negative sign

from the num numerator this gives us NE

* x^2 sin x + sin^2 x + 2x cosine x +

cosine 2 x all over x^2 + sin x^2

finally we can simplify this further by

using the identity sign 2 x + cosine 2 x

= 1 this gives us our final answer x^2

sin x + 2x cosine x + 1 all over x^2 +

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

successfully found the derivative using

the quotient rule the key steps were

identifying u and v finding their

derivatives applying the quotient rule

formula and then simplifying the result

Mastering Derivatives: Differentiating [(𝒄𝒐𝒔 𝒙)/(𝒙^𝟐+𝒔𝒊𝒏 𝒙)] Using the Quotient Rule

Unlock the secrets of calculus by learning how to differentiate the expression [(𝒄𝒐𝒔 𝒙)/(𝒙^𝟐+𝒔𝒊𝒏 𝒙)]  using the quotient rule. This video provides a step-by-step explanation, perfect for students and enthusiasts looking to deepen their understanding of derivatives involving trigonometric and polynomial functions.
#Calculus #QuotientRule #Trigonometry #MathTutorial #Differentiation


let's find the derivative of secant of a

x using first principles we'll break

this down step by step let's start by

defining our function let F ofx equal

secant of ax then F ofx + H equal secant

of a time x + H which equals secant of

X+ ah using the definition of the

derivative from first principles we know

that the derivative equals the limit as

H approaches Z of f of x + H - F ofx all

divided by H substituting our secant

function we get the limit as H

approaches zero of secant of ax plus a h

minus secant of ax all over H since cant

is 1 / cosine we can rewrite this as the

limit as H approaches Z of 1 / cosine of

a x + a h - 1 / cosine of a x all over H

to subtract these fractions we need a

common denominator this gives us cosine

of a xus cosine of a x + ah all over the

product of cosine of ax and cosine of ax

+ ah now we can use the cosine

difference formula this states that

cosine a minus cosine Bal -2 * s of a

plus b / 2 * s of a minus b/2

simplifying the Expressions inside the

sign functions we get 2 sin of a x + a h

over 2 * s of a h over 2 now we can

evaluate the limits separately as H

approaches zero the sign of a X Plus 8

h/2 approaches s of ax s of H h/2 over a

h two approaches 1 and cosine of x plus

h approaches cosine of a x this gives us

a Time sign of ax over cine of ax^2

which we can rewrite as a tangent of ax

* secant of ax therefore the derivative

of secant of ax equals a tangent of ax *

secant of a x and that's how we find the

derivative of secant of ax using first

principles

Understanding the Derivative of sec ax  Using First Principles

Master calculus fundamentals by deriving the exponential function sec(ax) from first principles. Use our step-by-step explanations and examples to deepen your understanding of derivatives and exponential functions.

#Calculus #Derivatives #ExponentialFunctions #MathTutorial #FirstPrinciples #x^3+2x


welcome today we'll learn how to find

the derivative of a complex exponential

function using logarithmic

differentiation our problem is to find

the derivative of y = x raed to the

cosine of x this is a challenging

function to differentiate directly

because we have a variable base and a

trigonometric function in the exponent

the strategy we'll use is logarithmic

differentiation which involves taking

the natural logarithm of both sides

first let's solve this step by step

First We Take the natural logarithm of

both sides using the logarithm property

that the natural log of a power equals

the exponent times the natural log of

the base we get Ln y = cine x * Ln X now

comes the key step we differentiate both

sides with respect to X on the left side

we get d by DX of Ln y on the right side

we need to use the product rule since we

have the product of two functions let's

handle the left side first remember that

the derivative of Ln y with respect to x

= 1/ y *

dydx this comes from the chain Rule now

let's evaluate the derivatives on the

right side we know that the derivative

of L l x is 1 /x and the derivative of

cosine X is sinx substituting these in

we get 1 / y * Dy DX = cosin x/ x + Ln x

* sinx to solve for dydx we multiply

both sides by y then we substitute back

y = x cosine X to get our final answer

and there we have it we've found the the

derivative using logarithmic

differentiation this technique is

particularly useful when dealing with

functions that have variables in both

the base and exponent the key steps we

used were one taking the natural

logarithm of both sides two using

logarithm properties three applying the

product rule for using the chain Rule

and finally substituting back the

original function

Mastering Logarithmic Differentiation: Solving [y = x^(cos x)] Step by Step

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

🔍 In this comprehensive tutorial, we break down how to find the derivative of y = x^(cos x) using logarithmic differentiation. Perfect for calculus students and math enthusiasts!
Key topics covered:
• Logarithmic differentiation technique
• Natural logarithm properties
• Product rule application
• Chain rule implementation
• Step-by-step solution process
💡 Learning Outcomes:
Master logarithmic differentiation
Handle complex exponential functions
Apply differentiation rules effectively
#Calculus #Mathematics #LogarithmicDifferentiation #MathTutorial #CalcTutorial #MathEduc


let's find the derivative of tangent 4X

using the first principle's definition

of derivatives let's start by defining

our function let F ofx = tangent 4X

therefore f of x + H = tangent of 4 * x

+ H now let's recall the definition of a

derivative the derivative of f ofx is

equal to the Limit as h approaches Z of

f ofx + H - F ofx all divid H let's

substitute our function into this

definition the derivative of tangent 4X

equals the limit as H approaches Z of

tangent of 4 * x + H - Tang 4X all

divided H now let's rewrite tangent in

terms of s over cosine this gives us the

limit as H approaches Z of s 4x + h/

cosine of 4x + H - sin 4X cosine 4X all

divided by H to subtract these fractions

we need a common denominator multiplying

both numerator and denominator of each

fraction by the denominator of the other

gives us this expression the numerator

is the S of the difference of angles

formula this simplifies to S of 48 H in

the numerator let's factor out the four

from the sign term this gives us 4 * the

limit as H approaches 0 of sin 4 H over

4 H divided by the product of the

cosiness now we can split this into two

limits the limit of sin 4 H over 4 H as

H approaches 0 = 1 and the limit of cine

4x + H as H approaches 0 = cosine 4X

this simplifies to 4 over cine 4X which

equal 4 secant 4X and that's our final

answer to verify this result we could

graph tangent 4X and it's derivative to

see that the derivative is indeed

steepest at the same points where

tangent 4X has its steepest slopes

Understanding the Derivative of tan4x  Using First Principles

Master the fundamentals of calculus by deriving the exponential function tan4x from first principles. Our step-by-step explanations and examples will enhance your grasp of derivatives and exponential functions.
#Calculus #Derivatives #ExponentialFunctions #MathTutorial #FirstPrinciples #tan3x


let's solve this calculus problem step

by step we need to find the derivative

of inverse tangent of square root of x

let's break this down into steps first

we'll use the chain rule since we have a

composite function remember the

derivative of inverse tangent of U is 1

1 + u^2 time the derivative of U in our

case U equal Square < TK of X so let's

substitute this into our chain rule

formula now we need to find the

derivative of square < TK of X using the

power rule with a fractional exponent we

know that the derivative of x to the 1/2

power is 12 * x -2 power which

simplifies to 1 / 2 < tkx let's

substitute this back into our equation

notice that in the denominator Square <

TK of X2 simplifies to just X finally we

can multiply these fractions together

remember when multiplying fractions we

multiply the numerators together and

multiply the denominators together this

gives us our Final Answer 1 / 2 < tkx *

1 + x let's verify this makes sense our

answer is a fraction that approaches

zero as X gets very large which is what

we would expect for the derivative of an

inverse tangent function since inverse

tangent approaches Pi / 2 as X

approaches infinity and that's how we

find the derivative of inverse tangent

of square < TK of X using the chain Rule

and power rule

Mastering Derivatives: Understanding the Derivative of [𝑡𝑎𝑛^(−1) √𝑥]

Dive into the world of calculus as we explore the step-by-step process of finding the derivative of  [𝑡𝑎𝑛^(−1) √𝑥]. Learn how to apply the chain rule effectively and simplify complex expressions with ease. Perfect for students looking to strengthen their understanding of mathematics concepts.
#Calculus #MathTutorial #Derivatives #Trigonometry #MathHelp #Education #STEM


welcome today we'll solve a challenging

derivative problem using the quotient

rule let's start by looking at our

function to solve this we'll use the

quotient rule which states that for a

function f ofx = u over V the derivative

is V * the derivative of U minus U * the

derivative of V all over V ^2 let's

identify our u and v terms u = e to x +

natural log of x and V equal log base of

X now let's find the derivatives of u

and v separately the derivative of U is

the derivative of e to the X which is e

to X plus the derivative of natural log

of x which is 1 /x for the derivative of

V we need to use the chain Rule and the

fact that log Bas of X = natural log of

x over natural log of a now let let's

substitute these into the quotient rule

formula in the numerator we multiply log

base a of x * the derivative of U minus

the quantity e to X Plus natural log of

x * the derivative of V the denominator

is log base uh of x^2 and this gives us

our final answer let's verify that this

matches our Target result the numerator

has log Basa of x * e x + 1 /x - e x +

natural log of x * log base a of e overx

all over log base a of x^2 let's

summarize the key steps we used to solve

this problem first we identified the

quotient rule pattern second we found

the derivatives of both the numerator

and denominator separately finally we

substituted everything into the quotient

rule formula and simplified this

technique can be applied to many complex

derivative problems involving quotients

Mastering Differentiation: Solving Complex Quotient Derivatives [((𝑒^𝑥+ln𝑥)/log_𝑎𝑥 )]

Unlock the secrets of differentiation with our detailed walkthrough on solving complex quotient derivatives. In this session, we’ll tackle a step-by-step solution to finding the derivative of [((𝑒^𝑥+ln𝑥)/log_𝑎𝑥 )]  using the quotient rule. Perfect for students and math enthusiasts looking to deepen their understanding of calculus concepts.
#Differentiation #Calculus #MathTutorial #QuotientRule #Derivatives #AdvancedMath


let's solve the step by step to find the

derivative of log base 10 of 3x first

let's write out what we're trying to

find we want to find the derivative with

respect to X of log base 10 of 3x the

first key step is to use the change of

Base formula this allows us to rewrite

our log base 10 in terms of the natural

logarithm which is easier to

differentiate the formula states that

log base 10 of 3x equals the natural log

of 3x * log base 10 of e next we can use

the fact that log base 10 of e is a

constant when we differentiate a

constant times a function we can factor

out the constant now we need to

differentiate the natural log of 3x

we'll use the chain rule here the

derivative of the natural log is 1/ its

input times the derivative of the input

the derivative of 3x is simply 3 so we

have log base 10 of e * 1 3x * 3 finally

we can simplify this expression the

threes cancel out leaving us with log

base 10 of e overx let's review what we

did we started with the derivative of

log base T of 3x we used the change of

Base formula to rewrite it in terms of

natural log we factored out the constant

log base 10 of E we use the chain rule

to differentiate the natural log and

finally we simplify to get our answer

log base 10 of e/x

How to Differentiate log_10(3𝑥)  Step by Step

Question: Find The Derivative of [ log_10(3𝑥)]

Learn how to differentiate the function  log_10(3𝑥)   using the change of base formula and chain rule. This step-by-step guide simplifies the process and helps you understand the underlying concepts. Perfect for math students wanting to improve their calculus skills!
#Calculus #Differentiation #MathTutorial #Logarithms #ChainRule


let's solve this complex derivative

problem involving inverse tangent

functions our goal is to find the

derivative of this expression let's

start by letting y equal our expression

we'll Begin by simplifying what's inside

the inverse tangent function our first

step is to factor be out of both the

numerator and

denominator notice how we can rewrite

the numerator as B * the quantity over

B+ X

and the denominator as B * the quantity

1 - over B * X since B appears in both

numerator and denominator we can cancel

it out this gives us a simpler form now

here's where we use a key Insight this

form matches a special inverse tangent

identity when we have an expression of

the form tangent inverse of U + V over

1us UV it equals tangent inverse of U

plus tangent inverse of V applying this

identity to our problem we can split

this into the sum of two inverse tangent

terms the first term will be tangent

inverse of A over B and the second term

will be tangent inverse of X now we can

differentiate the derivative of y with

respect 2x equals the derivative of

tangent inverse of over B plus the

derivative of tangent inverse of X let's

evaluate each term the first term is the

derivative of a constant so that's zero

for the second term we know the

derivative of inverse tangent of X is 1

1 + x^2 therefore our final answer is 1

over 1 + x^2 this elegant result shows

how a complex expression can be simplifi

to a basic derivative form through

careful algebraic manipulation and the

use of trigonometric identities

How to Differentiate [𝑡𝑎𝑛^(−1)((𝑎+𝑏𝑥)/(𝑏−𝑎𝑥))]  Step-by-Step

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

Unlock the secrets of calculus with this clear and concise tutorial on differentiating inverse trigonometric functions. Follow along as we break down the differentiation of [𝑡𝑎𝑛^(−1)((𝑎+𝑏𝑥)/(𝑏−𝑎𝑥))] . with step-by-step guidance. Perfect for students and math enthusiasts looking to deepen their understanding of trigonometric identities and differentiation techniques. Don’t forget to like, share, and subscribe for more math insights!
#Calculus #Trigonometry #MathTutorial #Differentiation #InverseTrigonometricFunctions #LearnMath


let's work through finding the

derivative of x^2 * e to x * natural log

of x this is a complex derivative that

will require multiple applications of

the product rule let's recall the

product rule first the derivative of U *

V equal U * the derivative of V + V *

the derivative of U for our function we

have three parts being multiplied

together x^2 e to X and natural log of x

let's break this down into two parts

first using the product rule first on

x^2 and G ofx we get x^2 * the

derivative of G of X Plus G ofx * the

derivative of x^2 now we need to find

the derivative of G of X which is e to x

* natural log of x using the product

rule again we get e to X time the

derivative of natural log of x plus

natural log of x * the derivative of e

to X let's substitute in the known

derivatives the derivative of natural

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

to X is e to the X the derivative of x^2

is 2x substituting these back into our

equation gives us x² * the quantity e to

X over x + natural log of x * e to X

+ 2x * e to x * natural log of x now

let's distribute and simplify the first

term becomes x e to x + x^2 * natural

log of x * e to x + 2x * e to x *

natural log of x finally we can factor

out x e to the X giving us our final

answer x e to x * the quantity 1 1 + x

natural log of x + 2 natural log of x

this problem demonstrated how to handle

a complex derivative using multiple

applications of the product rule along

with knowing our basic derivative rules

for exponential and logarithmic

functions the key was breaking down the

problem into manageable pieces and

applying the rules step by step

Finding the Derivative of (𝑥^2 𝑒^𝑥  ln𝑥)  Using the Product Rule | Step-by-Step Explanation.

In this video, learn to differentiate  (𝑥^2 𝑒^𝑥  ln𝑥) using the Product Rule in calculus.

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

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


let's find the derivative of sin 2x

using first principles this means we'll

use the fundamental definition of the

derivative to find the result let's

start by defining our function let F ofx

= sin 2x then f of x + H = sin of 2 *

the quantity x + H the derivative is

defined as the limit as H approaches

zero of f ofx + H -

FX all divid H now let's substitute our

function into the derivative definition

we get the limit as H approaches Z of s

of 2 * thean x + h- sin 2x all divid H

to solve this we'll use the

trigonometric identity for the

difference of signs the identity states

that S A minus sign b = 2 * cosine of a

plus b/ 2 * s of aus b/ 2 let's apply

this identity to our expression let's

simplify what's inside the cosine and S

2x + H + 2x = 4x + 2 h / 2 which = 2 x +

H and 2 x + H - 2x = 2 h / 2 which equal

H now we can split this into two parts

we have 2 * the limit as H approaches Z

of cosine of 2x + H times the limit as H

approaches Z of sin H H let's evaluate

each limit the limit of cosine of 2x + H

as H approaches 0 is just cosine of 2x

and the limmit of sign H H as H

approaches 0al 1 this is a fundamental

limit in calculus therefore the

derivative of sin 2x = 2 * cine 2x this

is our final answer this result shows us

that when we differentiate sin 2x we get

2 cine 2x the factor of two comes from

the chain rule which we applied

implicitly through our limit calculation

Understanding the Derivative of sin2x  Using First Principles

Dive into the fundamentals of calculus as we explore how to determine the derivative of the exponential function sin2x 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 #sin 2x


welcome today we'll learn how to find

the derivative of y = 1 + x raed to the

X power using logarithmic

differentiation this is a challenging

derivative that becomes much easier when

we use logarithms let's solve this step

by step first let's say y = 1 + x ra to

the X power now here's the key step we

take the natural log of both sides this

is what makes logarithmic

differentiation special using the

logarithm property that the log of a

power equals the power * the log of the

base we can simplify the right side to x

* Ln of 1 +

x now we can take the derivative of both

sides with respect to X on the left side

we use the chain rule the derivative of

n y is 1 y * Dy DX on the right side we

use the product rule since we have x *

Ln of 1 +

x let's simplify the right side the

derivative of Ln of 1 + x is 1 1 + x and

the derivative of x is just 1

now multiply both sides by y to isolate

dydx finally substitute back that y = 1

+ x to X

power and there we have it we've found

the derivative using logarithmic

differentiation this method is

particularly useful when dealing with

functions where one quantity is raised

to a variable power the key steps were

one taking the natural log of both sides

two using logarithm properties 3

differentiating both sides four using

the chain Rule and product rule five

solving for dydx

How to Find the Derivative of [(1+x)^x] | Logarithmic Differentiation Made Easy

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

📚 Master logarithmic differentiation with this step-by-step tutorial! Learn how to find the derivative of (1+x)^x using a powerful technique that simplifies complex exponential functions.
In this video, we cover:
✅ Understanding logarithmic differentiation
✅ Step-by-step solution process
✅ Application of chain rule and product rule
✅ Simplification techniques
✅ Final answer derivation
🔍 Perfect for:
• Calculus students
• Math enthusiasts
• Exam preparation
• Concept review
📝 Follow along as we break down each step:
Taking natural logarithms
Using log properties
Differentiating both sides
Applying derivative rules
Solving for dy/dx
⭐ Final Result: dy/dx = (1+x)^x[x/(1+x) + ln(1+x)]
Hashtags:
#Mathematics #Calculus #MathTutorial #Derivatives #LogarithmicDifferentiation #MathEducation #LearnMath #MathHelp


let's solve this challenging derivative

problem step by step we need to find the

derivative of y = x to X power * natural

log of x this is a complex function

involving both exponential and

logarithmic terms to solve this we'll

use logarithmic differentiation let's

start by taking the natural logarithm of

both sides using the properties of

logarithms we can

this into a sum the natural log of a

product equals the sum of the natural

logs so this becomes natural log of x to

the X Plus natural log of natural log X

now we can further simplify the first

term remember the natural log of x to a

power equals that power time the natural

log of x so this becomes x * natural log

X+ natural log of natural log X

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 when

differentiating natural log of Y on the

right side we need to use both the

product rule and chain rule for X Ln X

we get x * the derivative of Ln x + Ln x

* the derivative of x for Ln of Ln X we

get 1 Ln x * 1X let's simplify this we

get 1 + Ln x + 1 /x Ln X then multiply

both sides by y finding a common

denominator in the brackets we get X Ln

x + Ln x * X Ln x +1 all over X Ln X

finally we substitute back y = x x * Ln

X giving us our final answer the

derivative equal x x * Ln x * the

fraction 1 + x Ln x * 1 + Ln X all over

X Ln X this problem demonstrates the

power of logarithmic differentiation

when dealing with complex products and

exponential functions

Mastering Complex Derivatives Step-by-Step Solution for 𝑑/𝑑𝑥 [𝑥^𝑥  ln𝑥 ]

Question:
Find the derivative of 𝑑/𝑑𝑥 [𝑥^𝑥  ln𝑥 ] with respect to x.

🔷 In this comprehensive mathematics tutorial, we break down the solution for finding the derivative of [𝑥^𝑥  ln𝑥 ] using logarithmic differentiation.
📚 What you’ll learn:
• Logarithmic differentiation technique
• Using log properties effectively
• Applying chain rule and product rule
• Step-by-step problem solving approach
• Clear explanations for each mathematical step
💡 Perfect for:
• Calculus students
• Math majors
• Engineering students
• Test preparation
• Advanced mathematics enthusiasts
#Mathematics #Calculus #Derivatives #MathTutorial #LogarithmicDifferentiation #ChainRule #ProductRule #MathEducation #STEM #MathHelp


let's solve this calculus problem step

by step using the quotient rule we need

to find the derivative of the function f

ofx = x^2 + 1 / x^2 + 3 first let's

recall the quotient rule formula the

derivative of U over v = v * the

derivative of U minus U * the derivative

of V all over v^2 let's identify our u

and v function

U of X = x^2 + 1 and V of X = x^2 + 3

next we need to find the derivatives of

u and v the derivative of U equal 2X and

the derivative of V also equal 2x now

let's substitute everything into the

quotient rule formula we get x^2 + 3 *

2x - x^2 + 1 * 2x all over X 2 + 3^ 2

let's distribute the 2x in the numerator

in the first term we get 2x CU + 6x and

in the second term we get 2x Cub + 2x

now we can combine like terms in the

numerator the 2x cubed terms cancel out

and we're left with 6x - 2x which equal

4X therefore the derivative of x^2 + 1

/x^2 + 3 = 4X x^2 + 3^ 2 let's verify

this matches our original problem

looking back at our original function

and following the quotient rule steps

we've arrived at the correct derivative

Calculus Simplified: Derivatives of Rational Functions [(𝒙^𝟐+𝟏)/(𝒙^𝟐+𝟑)] Explained

Unlock the secrets of differentiating rational functions with this detailed tutorial. We walk you through the process of applying the quotient rule to  [(𝒙^𝟐+𝟏)/(𝒙^𝟐+𝟑)]  , ensuring you grasp each concept with clarity. Perfect for calculus students aiming to improve their understanding and skills!
#Calculus #Derivatives #QuotientRule #MathTutorial #LearnCalculus


today we'll solve an interesting

derivative problem involving inverse

trigonometric functions let's look at

the question before we dive into solving

this let's understand what we're dealing

with we have a composition of functions

First We Take the square root of x then

we take the sign of that result and

finally we take the inverse sign of

everything now here's a key Insight that

makes this problem much simpler when we

have sin inverse of sign of an angle

these operations cancel each other out

as long as the angle is within the

principle domain of inverse sign which

isk / 2 to < / 2 since Square < TK of X

is always positive it will always be in

the principal domain of inverse sign

therefore s inverse of s of sare < TK of

X simply equals < TK X now our problem

has become much simpler we just need to

find the derivative of square otk of X

let's solve this step by step first we

can rewrite square otk of X as x ^ 12

then we can use the power rule the

derivative of x n = n * x n -1 finally

we can rewrite x^ the -2 as 1/ < TK X

giving us our final answer 1 / 2 < tkx

and that's our solution the key to this

problem was recognizing that sin inverse

and sign cancel each other out within

the principal domain which greatly

simplified our calculation

Understanding Derivative [𝑠𝑖𝑛^(−1) (sin√𝑥 )] The Sine Inverse Function Explained

Dive into the world of calculus with our step-by-step breakdown of how to find the derivative of [𝑠𝑖𝑛^(−1) (sin√𝑥 )] . Learn the logic behind function simplification and apply the power rule effectively. Perfect for students and math enthusiasts aiming to strengthen their calculus skills. Watch now to enhance your understanding!
#Calculus #Derivatives #MathTutorial #SineInverse #LearningMath


welcome today we're going to learn about

logarithmic differentiation by finding

the derivative of a complex exponential

function specifically we'll find the

derivative of y = a to the power of a to

the power of X where it is a positive

constant why do we need logarithmic

differentiation here well this function

has a complex exponential structure that

would be very difficult to differentiate

directly by taking the natural logarithm

of both sides first we can transform

this into a simpler expression that's

easier to

differentiate let's solve the step by

step step one take the natural logarithm

of both sides this gives us Ln of Y = Ln

of a to the power of a to the power of X

step two use the logarithm property that

states Ln of a to a power equals that

power * Ln of a this gives us Ln of y =

a x * Ln of

a step three now we can differentiate

both sides with respect to X remember

when we differentiate Ln of Y I we get 1

/ y * Dy DX this is implicit

differentiation step four now we can

solve for

dydx multiply both sides by Y and since

y equal a to ^ of a to X we can

substitute this back in

let's talk about some common mistakes to

avoid first don't forget to use the

chain rule when differentiating a to the

X second remember that y equals your

original function when substituting back

and third be careful with your logarithm

properties let's verify our answer makes

sense the derivative should get larger

as X increases because our original

function grows very rapidly our answer

shows this because both a to the power

of a to the X and A to the X are

increasing functions here are some

practice tips for similar problems first

whenever you see complex exponentials

consider logarithmic

differentiation second break down the

problem into small steps and third

always verify your answer makes sense in

conclusion

we found that the derivative of y = a

power of a to X is a the power of a to

the x * a to x * Ln of a squar we did

this using logarithmic

differentiation which simplified our

complex exponential into a more

manageable form thank you for watching

Mastering Logarithmic Differentiation: Solving Complex Exponential Derivatives | d/dx[𝑎^(𝑎^𝑥 )]

QUESTION: Find the derivative of y = d/dx[𝑎^(𝑎^𝑥 )] using logarithmic differentiation.

📚 In this comprehensive calculus tutorial, we break down how to find the derivative of a(ax) using logarithmic differentiation. Perfect for AP Calculus, IB Mathematics, and college-level calculus students!
🎯 What You’ll Learn:
• Why logarithmic differentiation is the key to solving complex exponentials
• Step-by-step solution process
• Important logarithm properties and chain rule applications
• Common pitfalls to avoid
• Tips for tackling similar problems
📝 Prerequisites:
Basic differentiation rules
Natural logarithms
Chain rule
Exponential functions
✅ Download practice problems and solutions in the description below!
#Calculus #Mathematics #Derivatives #LogarithmicDifferentiation #MathTutorial #APCalculus #IBMaths #CalcTutoring #MathEducation #OnlineLearning


let's solve this complex derivative

problem step by step we need to find the

derivative of the natural logarithm of a

fraction involving square roots the

first step is to use the logarithm

property that states the log of a

fraction equals the log of the numerator

minus the log of the denominator this

allows us to split our expression into a

difference of two logarithms next we'll

use the chain rule to differentiate each

logarithm term remember the derivative

of natural log is 1 over the function

time the derivative of the function

let's apply this to both terms now we

need to find the derivative of square <

TK of x + 1 remember the derivative of

square < TK X is 1 / 2 * < TK X so the

derivative of square < TK of x + 1 will

be 1 / 2 < TK x + 1 * the derivative of

x + 1 the derivative of x + 1 is simply

1 so we can replace those terms with one

and begin to simplify our expression to

combine these fractions we need a common

denominator after combining and

simplifying the numerator we get this

expression looking at the numerator we

can see that the square root terms

cancel out leaving us with just two and

and in the denominator we can simplify x

+ 1 -1 to just X finally we can cancel

the twos and simplify the denominator to

get our final answer 1X * < TK x + 1

this problem demonstrated several

important calculus Concepts the

properties of logarithms the chain Rule

and the derivative of square root

functions by breaking it down step by

step we were able to solve this complex

derivative

Mastering Calculus: Differentiating Logarithmic Functions [𝒍𝒏((√(𝒙+𝟏)−𝟏)/(√(𝒙+𝟏)+𝟏))]

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

In this video, we break down the process of differentiating logarithmic functions step-by-step. Learn how to handle complex expressions involving square roots and logarithms using the chain rule. Perfect for students looking to enhance their calculus skills and understanding.

#Calculus #Differentiation #Logarithms #MathTutorial #Learning


let's solve this calculus problem step

by step we need to find the derivative

of secant inverse of X sared let's break

this down into manageable steps using

the chain rule remember when we have a

function raised to a power we can use

the chain rule which states that the

derivative of u^2 equals 2 * the

derivative of U in our case U equal

secant inverse of X so applying the

chain rule we get 2 * secant inverse of

x * the derivative of secant inverse of

X now we need to know the derivative of

secant inverse of X this is a standard

derivative that we should memorize it =

1 /x * < TK x^2 - 1 substituting this

back into our equation we get 2 * secant

inverse x * 1 /x * < TK of x^2 -1

finally we can write this as 2 * secant

inverse x/ x * < TK of x^2 - 1 this is

our final answer let's verify the key

components of this solution first we use

the chain rule correctly to handle the

squared term second we correctly applied

the derivative of inverse secant and

finally we simply simplified our

expression to its most compact form

remember when dealing with inverse

trigonometric functions it's crucial to

know their derivatives the derivative of

inverse secant is particularly useful in

calculus problems like this one

Master the Derivative of Inverse Secant Squared [(𝑠𝑒𝑐^(−1) 𝑥)^2] Step-by-Step Guide

Question:
Calculate the derivative of  [(𝑠𝑒𝑐^(−1) 𝑥)^2]   with respect to x .
Unlock the mystery of calculus with our detailed breakdown of the derivative of  [(𝑠𝑒𝑐^(−1) 𝑥)^2] . This step-by-step guide takes you through each mathematical operation and its explanation, helping you master this important calculus concept. Perfect for students, teachers, and anyone keen on understanding advanced math. Don’t forget to subscribe for more insightful math tutorials!
#Calculus #MathTutorial #Derivatives #InverseSecant #MathHelp


let's solve this calculus problem step

by step using the quotient rule we need

to find the derivative of sinx / by x^2

+ cosine X first let's recall the

quotient rule the derivative of U over V

equal V * the derivative of uus U * the

derivative of V all over V ^2 let's

identify our u and v functions U = sin x

and V = x² + cin X now let's find the

derivatives of u and v the derivative of

sinx is cine X the derivative of x^2 +

cine X is 2x - sin x let's substitute

these into the quotient rule formula we

get V which is x^2 + cosine x * the

derivative of U which is cosine x minus

U which is sin x * the derivative of V

which is 2x - sin x all over V ^2 now

let's expand the numerator first multip

x^2 + cosine X by cosine X this gives us

x^2 cosine x + cosine 2 x then multiply

sin x by 2x - sin x this gives us 2x sin

x - sin^2 x let's combine these terms in

the numerator we get x^2 cosine x +

cosine 2 x - 2x sin x + sin^2 x all over

x^2 + cosine x^2 finally we can simplify

this further remember that sin^2 x +

cosine 2 x = 1 using this our final

answer is x^2 cine x - 2x sin x + 1 all

over x^2 + cine x^2 and that's our

solution we used the quotient rule to

find the derivative expanded the terms

and simplified using trigonometric

identities to get our final answer

Differentiating [(𝒔𝒊𝒏 𝒙)/(𝒙^𝟐+𝒄𝒐𝒔 𝒙)] A Step-by-Step Guide Using the Quotient Rule

In this comprehensive calculus tutorial, learn how to differentiate the function [(𝒔𝒊𝒏 𝒙)/(𝒙^𝟐+𝒄𝒐𝒔 𝒙)] using the quotient rule. Follow along with clear, step-by-step explanations and tips to master this essential calculus technique. Perfect for students, educators, and anyone looking to enhance their mathematics skills.
#Calculus #Differentiation #QuotientRule #MathTutorial #Trigonometry


welcome today we'll solve a challenging

differentiation problem involving

inverse trigonometric functions let's

look at our question let's solve the

step by step first let's let y equal our

function now this expression inside the

ark sign function looks complicated to

simplify it let's make a clever

substitution let's say x = tan Theta

this means Theta equals inverse tangent

of X when we make the

substitution our expression becomes sin

inverse of 1 - tan^ 2 thet over 1 + tan^

2 thet now here's where it gets

interesting this fraction is actually a

well-known trigonometric identity it

equals cosine of 2

Theta we can simplify this further using

another identity Co sin of 2 thet = sin

ofun / 2 - 2 thet or equivalently sin of

piun / 2 + 2 Theta now we have sin

inverse of s these are inverse functions

so they cancel out leaving us with Just

pi/ 2 + or minus 2 Theta remember that

Theta is inverse tangent of X so we can

write our final expression for y as Pi /

2 + or - 2 inverse tangent of X now we

can differentiate this expression with

respect to X the derivative of a

constant is zero and the derivative of

inverse tangent is 1 1 + x^2 therefore

the derivative of sin inverse of 1 - x^2

over 1 + x^2 = + or - 2 over 1 + x^2

this was a challenging problem that

required several key insights making a

clever substitution recognizing

trigonometric identities and carefully

applying the chain rule thank you for

watching

Differentiating Inverse Trigonometric Functions [𝑠𝑖𝑛^(−1)((1−𝑥^2)/(1+𝑥^2 ))] A Step-by-Step Guide

Question: Differentiate the function [𝑠𝑖𝑛^(−1)((1−𝑥^2)/(1+𝑥^2 ))]   with respect to x.

Learn how to differentiate the function  [𝑠𝑖𝑛^(−1)((1−𝑥^2)/(1+𝑥^2 ))] using trigonometric identities and calculus techniques. This detailed walkthrough will help you understand the process step-by-step with clear explanations.
#Calculus #Trigonometry #MathTutorial #Derivative #MathHelp


let's solve this calculus problem step

by step we need to find the derivative

of the cube root of x * sinx first let's

identify our two functions let's call U

of X = the cubot of X which we can write

as x^ the 1/3 power and V ofx = sinx now

since we're multiplying these functions

together 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 let's

find each derivative separately the

derivative of sin x is cine X and the

derivative of x^ 1/3 power is 1/3 * X

-23 power now let's substitute

everything into the product rule we get

x 1/3 * cosine X Plus sin x * 1/3 x

-23 let's simplify the second term the

fraction 1/3 and x to the -2/3 can be

Rewritten as sin x over 3x^ 2/3 power

finally we can factor out x to the 1/3

power our final answer is x 13 * the

quantity cosine X+ sin x over 3x let's

review what we did we identified our

functions and Ed the product rule found

the derivatives of each function

separately and then simplified our

expression to get our final answer

Finding the Derivative of ∛𝑥  sin𝑥  Using the Product Rule | Step-by-Step Explanation

In this video, learn to differentiate ∛𝑥  sin𝑥  using the Product Rule in calculus.

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

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


let's find the derivative of x Cub + 2x

using the definition of the derivative

first let's write our function f ofx = x

Cub + 2x the definition of the

derivative states that it's equal to the

Limit as H approaches Z of f ofx + hus F

ofx all divided H step one let's find f

of x + H this means we replace every X

in our original function with x + H when

we expand this we get x + H cubed + 2 *

x + H step two let's expand x + H cubed

using the cube expansion formula we get

X cubed +

3x^2 H + 3x h^2 + H cubed and don't

forget to add 2x + 2 H from the second

term step three now we can substitute

this into our derivative definition we

subtract F ofx which is X Cub + 2x and

divide everything by H step four let's

simplify the numerator the X cubed terms

and 2x terms cancel out this leaves us

with 3x^2 H + 3x h^2 + HB + 2 H all over

h step five we can factor out H from the

numerator this gives us H * the 3x^2 +

3xh + h^2 + 2 all over H the H's cancel

out step 6 now we can evaluate the limit

as H approaches zero the terms with h go

to zero leaving us with 3 x^2 + 2

therefore the derivative of x Cub + 2x

is 3x^2 + 2 this is a great example of

using the definition of the derivative

to find the derivative of a polom

function while we could have used the

power rule understanding the definition

helps us grasp what a derivative really

means

Understanding the Derivative of x^3+2x  Using First Principles

Master calculus fundamentals by deriving the exponential function x^3 + 2x from first principles. Utilize our step-by-step explanations and examples to strengthen your understanding of derivatives and exponential functions.

#Calculus #Derivatives #ExponentialFunctions #MathTutorial #FirstPrinciples #x^3+2x


welcome to this calculus lesson on

logarithmic differentiation today we'll

learn how to find the derivative of a

complex expression sin x raed to the

cosin X Plus cine x raed to the sinx

let's break this down into manageable

steps first let's separate this into two

parts we'll call the first part you and

the second part five this will help us

handle each EXP ential terms separately

now let's focus on finding dudx first

this is where logarithmic

differentiation comes in handy we'll

take the natural log of both sides

remember when we take the natural log of

an exponential expression we can bring

the exponent down as a coefficient now

we can differentiate both sides with

respect to X remember we'll need to use

the chain rule here on the left side

we get 1 / U * Dux on the right side we

need to use the product rule since we

have two functions being multiplied

together let's solve for DX we multiply

both sides by U which is sin x to the

cosine X now let's find dvdx using the

same process we take the natural log of

V which gives us sin x * Ln of cine X

again we differentiate both sides using

the same process as before we get 1 / V

* dvdx equals the derivative of sin x *

Ln of cosine x + sin x * the derivative

of Ln of cosine X solving for dvdx we

multiply both sides by V which is cosine

X to the sinx finally we can combine our

results the derivative of y equal DX x +

dvdx this gives us our final answer sin

x cosine x * the quantity negative sin x

Ln sin x + c^ 2 x/ sin x plus cosine X

sin x * the qu cosine X Ln cosine x -

sin^2 XC X let's review some common

mistakes to avoid one don't forget to

use the product rule when

differentiating terms with natural logs

two remember to multiply by the original

function when solving for the derivative

three be careful with your signs when

dealing with negative terms four don't

forget to combine both parts at the end

and that's how we use logarithmic

differentiation to find the derivative

of this complex expression thanks for

watching

Master Logarithmic Differentiation: Solving (sin x)^(cos x) + (cos x)^(sin x) | Calculus Made Clear

Question: Find the derivative of (sin x)^(cos x) + (cos x)^(sin x)  with respect to x.

📚 In this comprehensive calculus tutorial, we break down a challenging derivative problem using logarithmic differentiation. Perfect for university calculus students and math enthusiasts!
🎯 What You’ll Learn:
• How to tackle complex exponential functions
• Step-by-step logarithmic differentiation technique
• Mastering the chain rule and product rule
• Working with trigonometric functions
• Common pitfalls and how to avoid them
💡 Prerequisites:
• Basic differentiation rules
• Understanding of natural logarithms
• Knowledge of trigonometric functions
✏ Practice problems and solutions available in the description below!
📝 Related Resources:
#Calculus #Mathematics #LogarithmicDifferentiation #MathTutorial #CalcTutorial #MathEducation #STEM #LearnMath


let's work through finding the

derivative of e to the power of the

square otk of X step by step our

original function is y = e power of the

square < TK of x to solve this we'll

need to use the chain rule remember the

chain rule states that when we have a

composite function we multiply the

derivative of the outer function by the

derivative of the inner function in our

case the outer function is e to the

power of something and the inner

function is the square root of x let's

start with step one the derivative of e

to the power of U with respect to U is

just e to the power of U this is our

outer function derivative step two we

need to find the derivative of the inner

function which is the derivative of

square < TK of X the derivative of

square < TK of X is 1 / 2 root of x step

three now we apply the chain rule by

multiplying these parts together we

multiply e to power of s < TK x * 1 / 2

< TK X our final answer can be written

as E power of s < TK x/ 2 < TK X let's

visualize why this makes sense here's

the graph of our original function e to

the power of square < TK X and here's

its derivative that we just found notice

how the derivative shows the rate of

change of our original function at each

point and that's how we find the

derivative of e to the power of square

otk of X using the chain rule remember

when dealing with composite functions

like this always break it down into

steps using the chain Rule and multiply

the derivatives of the outer and inner

functions together

9.3-2.1 Mastering the Chain Rule: Differentiating (𝑒^√𝑥)  Step-by-Step

Unlock the secrets of differentiation with our in-depth tutorial on finding the derivative of (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


let's solve this differentiation problem

step by step we need to differentiate

the expression x^2 + 1 * inverse tangent

of x - x with respect to X let's break

this down into manageable steps first

we'll use the product rule on the first

term since it's a product of two

functions and then separately

differentiate the minus X term now let's

evaluate each part the derivative of

inverse tangent is 1/ 1 + x^2 the

derivative of x^2 + 1 is

2X and the derivative of x is simply 1

substituting these derivatives back into

our expression we get x^2 + 1 * 1 1 +

x^2 plus inverse tangent x * 2x - 1

Let's simplify the first term notice

that x^2 + 1 over 1 + x^2 simplifies to

1 since these terms are identical

finally we can combine like terms 1 -

one cancels out leaving us with our

final answer 2x * inverse tangent X to

verify this answer you could take the

derivative of our result and check if it

matches the original function or verify

that this solution satisfies any given

initial conditions

Understanding the Differentiation of [(𝑥^2+1) 𝑡𝑎𝑛^(−1)  𝑥−𝑥]

Question: Differentiate [(𝑥^2+1) 𝑡𝑎𝑛^(−1)  𝑥−𝑥]  with respect to x.
In this video, we explore the step-by-step differentiation of [(𝑥^2+1) 𝑡𝑎𝑛^(−1)  𝑥−𝑥]. Watch as we break down the process using the product rule and basic differentiation principles to arrive at the simplified expression. Perfect for students and math enthusiasts looking to deepen their understanding of calculus!

#Calculus #Differentiation #MathTutorial #ProductRule #ArctanDifferentiation


let's solve this derivative step by step

we'll start with the original expression

and simplify it before taking the

derivative first let's recall the double

angle formula for cosine cine of 2x = c^

2 x - sin^2 x we'll substitute this into

our expression now let's distribute the

negative sign in the numerator this

gives us cosine x - cos sin^2 x + s^ 2

x/ 1 - cosine X here's where we can use

another important trigonometric identity

remember that sin^2 X+ cine 2 x = 1 so

sin^2 x = 1 - cosine 2 x let's

substitute this into our expression now

we have cine x - c^ 2 X x + 1 - c^ 2 X

all over 1 - cine X let's group terms in

the numerator first we can factor out 1

- cosine X from the first two terms then

we can Factor the remaining terms with 1

- cine x * 1 + cine X now we can cancel

the 1 - cine X terms in the numerator

and denominator this simplifies our EXP

expression to 2 cine x + 1 now that we

have simplified our expression we can

easily take the derivative the

derivative of cine X is sinx and the

derivative of a constant is zero

therefore our final answer is -2 sinx

this completes our solution to visualize

this here's what the original function

and its derivative look like graphically

Simplifying Trigonometric Expressions [(cos𝑥−cos2𝑥)/(1−cos𝑥)] Calculus with Cosine Functions

Join us as we break down the process of simplifying and differentiating complex trigonometric expressions. Learn how the expression  [(cos𝑥−cos2𝑥)/(1−cos𝑥)]   simplifies to (2cosx+)1  and its differentiation to (-2sinx) . Perfect for students aiming to conquer trigonometric identities and calculus differentiation techniques.
#Trigonometry #Calculus #MathTutorial #Differentiation #LearnMath


let's solve this calculus problem step

by step using the quotient rule we need

to find the derivative of 1+ sinx / 1 +

cosine X first let's recall the quotient

rule the derivative of U over v = v *

the Der of uus U * the derivative of V

all over V ^2 let's identify our u and v

functions U of x = 1 + sin x and

V ofx = 1 + cosine X next we need to

find the derivatives of u and v the

derivative of U is cine X and the

derivative of V is sinx now let's

substitute these into the quotient rule

formula we get 1 + cosine x * cine X - 1

+ sin x * sinx all over 1 + cine x^ 2

let's distribute in the numerator on the

left side we get cine x + cine 2 X on

the right side we get sin x + sin^2 x

combining these terms in the numerator

we get cosine x + cosine 2 x + sin x +

sin 2 x all over 1 + cosine x² now

remember that sin^2 x + cosine 2 x = 1

using this identity we can simplify our

numerator to 1 + sin x + cine X and this

is our final answer we've successfully

found the derivative using the quotient

rule to verify this is correct we can

check that one we use the quotient rule

properly two we found the derivatives of

u and v correctly and three we simp

ified our trigonometric Expressions

correctly using the fundamental identity

all these steps check out so our answer

is correct

Master the Quotient Rule: Differentiating [(𝟏+𝒔𝒊𝒏𝒙)/(𝟏+𝒄𝒐𝒔𝒙)]  Explained

Join us in this engaging math tutorial where we break down the differentiation of the function [(𝟏+𝒔𝒊𝒏𝒙)/(𝟏+𝒄𝒐𝒔𝒙)]  using the powerful quotient rule. Perfect for students, teachers, and math enthusiasts, this video provides a step-by-step guide to understanding and applying calculus principles with ease. Strengthen your calculus skills and gain confidence with each problem solved.
#Calculus #QuotientRule #MathTutorial #Differentiation #Derivatives