let's solve this calculus problem step

by step using the quotient rule we need

to find the derivative of the function f

ofx = the < TK x + 1 over the < TK of x

-1 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 here = <

TK of x + 1 and V = sare < TK of x -1

next we need to find the derivatives of

u and v the derivative of square otk of

X is 1 / 2 < TK X so both do DX and dvdx

equal 1 / 2 s < TK of X now let's

substitute everything into the quotient

rule formula we get square < TK of x -1

* 1 2 < TK x - < TK of x +1 * 1 2 < TK X

all over < TK x - 1^ 2 let's simplify

the numerator first we can factor out 1/

2 < TK X from both terms then inside the

parentheses we have S < TK of x - 1 - <

TK of x + 1

simplify further the square root of x

terms cancel out and we're left with -2

in the brackets so our numerator becomes

-1/ < TK X and finally our answer is -1

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

solution demonstrates the power of the

quotient rule in calculating derivatives

of complex fractions remember to

carefully identify your u and v

functions find their derivatives and

then apply the quotient rule formula

step by step

Mastering the Quotient Rule: Differentiating Fractions ((√𝒙+𝟏)/(√𝒙−𝟏)) Involving Radicals

Learn how to differentiate complex rational functions involving radicals using the quotient rule. In this tutorial, we’ll guide you through each step of finding the derivative of ((√𝒙+𝟏)/(√𝒙−𝟏)). Perfect for students looking to excel in calculus, this video makes differentiation simple and accessible!
#Calculus #QuotientRule #Differentiation #Radicals #MathHelp


let's solve this challenging derivative

problem step by step we need to

differentiate sin squar of natural log

of x^2 with respect to X this is a

complex function with multiple nested

parts let's break it down using the

chain rule we have sin squar as our

outer function then natural log of x^2

as our inner function let's apply the

chain rule step by step first the

derivative of sin s of U is 2 sin of U *

the derivative of s of U next the

derivative of s of U is cosine of U *

the derivative of U so our expression

becomes 2 sin of natural log of x^2 *

cosine of natural log of x^2 * the

derivative of natural log of

X2 now the derivative of natural log of

x

is 1 /x^2 * the derivative of

x^2 the derivative of x^2 is 2x so our

expression becomes 2 sin of natural log

of x^2 * cosine of natural log of x^2 *

1x^2 *

2x simplifying this expression we get

four sign of natural log of x^2 * cosine

of natural log of x^2

/x now we can use the double angle

formula for S which states that 2 sin of

U * cosine of U equal s of 2

you applying this formula to our

expression we get 2 sin of 2 natural log

of x^2 /x

finally we can simplify this further by

noting that 2 natural log of x^2 = 4

natural log of x giving us our final

answer 2 sin of 4 natural log of

XX and there we have it we've

successfully differentiated this complex

function using the chain Rule and

trigonometric identities remember to

break down complex problems into smaller

manageable steps

Calculus Made Easy: Differentiating [𝒔𝒊𝒏^𝟐 {𝒍𝒏(𝒙^𝟐)}]  Step-by-Step

Unlock the secrets of calculus with our detailed walkthrough on differentiating the expression [𝒔𝒊𝒏^𝟐 {𝒍𝒏(𝒙^𝟐)}] . This video provides a comprehensive, step-by-step explanation, perfect for students and enthusiasts looking to strengthen their understanding of calculus concepts. Learn how to apply the chain rule effectively and simplify complex derivatives with ease.
#Calculus #MathTutorial #Derivatives #ChainRule #MathHelp


let's solve this calculus problem step

by step we need to find the derivative

of a to the power of x * x^2 with

respect to X this is a product of two

functions so we'll need to use the

product rule the product rule states

that the derivative of U * V equal U *

the derivative of V plus v * the

derivative of U let's identify our u and

v in this case u = a the X and B = x^2

now let's find each derivative

separately first the derivative of x^2

is 2X and the derivative of a to X is a

to x * the natural log of let's

substitute these into the product rule

we get a to x * 2x + x^2 * a to x *

natural log of a now let's factor out

the common terms we can factor out a to

the X giving us a to x * thean 2x + x^2

* natural log of a for our final step we

can factor out x^2 from inside the

parentheses this gives us a to the x *

x^2 * 2 /x + natural log of a and this

is our final answer let's review what we

did we used the product rule because we

were different differentiating a product

of functions we found the derivatives of

each part separately and then we

factored our expression to get it into

its simplest form here's our complete

solution one more time starting from a

to the power of x * x^ 2 we use the

product rule and ended up with a to the

power of x * x^2 * 2/x + natural log of

a

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

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

What You'll Learn:
- Definition of the Product Rule
- Step-by-step differentiation of  (a^x .x^2)
- 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 solve this calculus problem step

by step using the quotient rule we need

to find the derivative of cosine Theta

minus sin Theta over cosine theta plus

sin Theta 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 sared before

we apply the quotient rule directly we

can simplify our expression

if we divide both numerator and

denominator by cine Theta we get 1us

tangent Theta over 1 + tangent Theta now

we can identify our u and v u = 1 - Tang

Theta and V = 1 + tangent Theta next we

need to find the derivatives of u and v

the derivative of U is secant thet and

the derivative of V is sec 2 th now

let's substitute everything into the

quotient rule formula we get 1 + tangent

th * SEC 2 thet - 1 - Tang thet * SEC

2et all over 1 + Tang Theta 2 let's

distribute in the numerator for the

first term we get secant 2 thetus secant

2 thet * tangent Theta for the second

term we get secant 2 Theta + secant 2

Theta * tangent Theta when we combine

like terms the tangent Theta terms

cancel out and we're left with -2 secant

Theta in the numerator and that's our

final answer the derivative of cine

Theta minus sin Theta over cine Theta +

sin Theta = -2 SEC 2 Theta over 1 +

tangent Theta 2 this problem

demonstrates the power of the quotient

rule combined with trigonometric

identities remember to always look for

ways to simplify your expression before

applying the quotient rule

Trigonometric Differentiation: Unraveling Complex Expressions [(𝒄𝒐𝒔  𝜽−𝒔𝒊𝒏 𝜽)/(𝒄𝒐𝒔  𝜽+𝒔𝒊𝒏 𝜽)]

In this video, we delve into differentiating complex trigonometric expressions. Learn how to differentiate [(𝒄𝒐𝒔  𝜽−𝒔𝒊𝒏 𝜽)/(𝒄𝒐𝒔  𝜽+𝒔𝒊𝒏 𝜽)] and simplify it to (( −𝟐𝒔𝒆𝒄^𝟐 𝜽 )  )/(𝟏+𝒕𝒂𝒏 𝜽)^𝟐 . Ideal for students aiming to master calculus and trigonometric functions.
#Trigonometry #Calculus #Differentiation #MathHelp #AdvancedMath


let's learn how to differentiate a

square root function containing a

quadratic expression we'll break this

down step by step to make it clear and

intuitive let's start with our function

we want to find the derivative of the

square root of a x^2 + BX + C this is a

complex function that we'll need to

handle carefully using the chain rule to

solve this we'll use the chain rule 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 the square

root and the inner function is the

quadratic expression ax^2 + bx+ C let's

handle the outer function first the

derivative of the square root function

with respect to U is 1 over 2 * U to the

-2 power we can rewrite this as 1/ 2 *

the square < TK of U now for the inner

function the derivative of ax^2 + BX + C

is simply 2 ax + B now let's combine

these parts using the chain rule we take

the derivative of the outer function

replacing you with our inner function

and multiply it by the derivative of the

inner function finally we can write our

answer in its simplest form the D

derivative equal 2 a x + B / 2 * the <

TK of ax^2 + BX + C let's verify our

work we use the chain rule correctly by

multiplying the derivative of the square

root function by the derivative of what

was inside it our final answer maintains

the proper form and includes all terms

from both the numerator and denominator

and that's how we differentiate a square

root function function containing a

quadratic expression remember to always

apply the chain rule carefully and keep

track of both the outer and inner

functions

9.3-1.3 Deriving the Square Root of Quadratic Functions (√(𝑎𝑥² + 𝑏𝑥 + 𝑐)): A Comprehensive Guide.

Dive into the fundamentals of calculus with our detailed tutorial on differentiating functions involving square roots. We’ll break down the process step-by-step using the chain rule, enhancing your understanding of how to handle complex derivatives. Perfect for calculus students and anyone looking to solidify their calculus skills.
#Calculus #Differentiation #SquareRoots #ChainRule #MathTutorial #cpmrana


today we'll solve a challenging

differentiation problem involving

inverse trigonometric functions let's

look at the question let's solve the

step by step first recall that when

differentiating inverse tangent we use

the formula the derivative of AR tangent

of u = 1 1 + u^2 time the derivative of

U in our case U = cosine x/ 1 + sin x

apply this formula we get 1/ 1 + u^2

where U is cosine X over 1 + sin x all

sared this can be Rewritten as 1/ 1 +

cine 2 x overan 1 + sin x^2 now for the

second part we need to differentiate

cine X over 1 + sinx this requires the

quotient rule remember the quotient rule

states that the derivative of f over G

equal G * the derivative of f - f * the

derivative of G all over G ^2 combining

these parts and simplifying the

denominator we get this fraction now we

can use the trigonometric identity sign

2 x + cosine 2 x = one to simplify

further after applying this identity and

combining like terms we get sinx - 1 / 2

sin x + 2 we can factor out -1 from the

numerator and two from the denominator

giving us -1 over2 * sin x + 1 / sin x +

1 the sin x + 1 terms cancel out leaving

us with our final answer -2 and there we

have it we've successfully

differentiated this complex inverse

Trigon ometric expression using the

chain rule quotient Rule and

trigonometric identities to arrive at

our surprisingly simple answer of - one2

Derivative of Arctan: A Step-by-Step Trigonometric Solution [𝑡𝑎𝑛^(−1) (cos𝑥/(1+sin𝑥 ))]

Question: Differentiate [𝑡𝑎𝑛^(−1) (cos𝑥/(1+sin𝑥 ))]  with respect to x .

In this video, we explore the process of differentiating the function  [𝑡𝑎𝑛^(−1) (cos𝑥/(1+sin𝑥 ))] step by step. We’ll apply various calculus techniques, including the derivative of inverse trigonometric functions, the quotient rule, and key trigonometric identities. By the end, you’ll understand how to break down complex expressions into simpler components. Perfect for calculus students and trigonometry enthusiasts!
 #Calculus #Trigonometry #Derivative #InverseTrigonometricFunctions #MathTutorial


welcome today we'll solve a challenging

differentiation problem involving an

inverse sign function with a complex

argument let's start with our problem we

need to find the derivative of sin

inverse of the square < TK of x * e to X

let's break this down step by step first

recall the derivative formula for

inverse sign the derivative of inverse

sign of U equal 1/ theare < TK of 1 - U

^2 time the derivative of U in our case

U equal the sare < TK of x e to X let's

substitute this into our

formula now let's simplify what's under

the first square root notice that when

we square the square root of x e to the

X we just get x e to

X next let's focus on the derivative of

square < TK of x e to X we can use the

chain rule here the derivative of square

otk of U is 1 / 2 < TK U time the

derivative of

U now we need to find the derivative of

x e to X this requires the product rule

the derivative will be x * the

derivative of e to X Plus e to x * the

derivative of x

X let's substitute this back into our

equation finally we can write this as a

single fraction our final answer is x e

to x + e to X all over 2 * the < TK of x

e to x * the sare < TK of 1 - x e to the

X let's verify this is correct by

checking our step steps we use the

inverse sign derivative formula apply

the chain rule correctly use the product

rule for x e to the X and simplified our

expression all steps follow standard

calculus rules so our answer is correct

and that completes our solution remember

when dealing with complex derivatives

like this breaking them down into

smaller manageable steps using standard

derivative rules makes the problem much

easier to solve thank you for watching

Mastering Derivatives: Differentiating Inverse Trigonometric Functions [𝑠𝑖𝑛^(−1) (√(𝑥𝑒^𝑥 ))]

Problem Statement
Differentiate the function [𝑠𝑖𝑛^(−1) (√(𝑥𝑒^𝑥 ))]  with respect to x.
Unlock the secrets to differentiating inverse trigonometric functions with this comprehensive step-by-step guide! In this video, we break down the process of finding the derivative of [𝑠𝑖𝑛^(−1) (√(𝑥𝑒^𝑥 ))] using chain and product rules, explained clearly and concisely. Perfect for students and math enthusiasts looking to deepen their calculus skills. Don’t miss out on understanding every detail with our two-column explanation method!
#Calculus #Derivatives #MathTutorial #InverseTrigonometry #ChainRule #ProductRule #MathHelp #LearnMath #StepByStep


let's solve this complex derivative step

by step we're going to differentiate sin

cubed of X cubed using the chain rule

let's start with our original expression

we need to find the derivative of sin

Cub of X cubed we can rewrite this as s

of X cubed all cubed to make it clear

what we're dealing with now let's apply

the chain rule first we'll use the power

rule on the outer function the

derivative of U cubed is 3 U ^2 in our

case U is s of X Cub so we get 3 * sin

of x cu^ 2 times the derivative of s of

X cubed next we need to find the

derivative of s of X cubed the

derivative of s of X with respect to X

is cosine of x but since we have X cubed

inside we need to multiply by the

derivative of the inner function now for

the final step we need need to find the

derivative of x cubed using the power

rule the derivative of x cubed is 3x

sared let's substitute this in finally

let's simplify our Expression 3 * 3 is 9

and we can rearrange the terms to get

our final answer 9 x^2 * sin^2 X Cub *

cosine of x cubed let's review what we

did we used the chain rule twice once

for the cube power and once for the sign

function we then used the power rule to

differentiate X cubed and finally

combined all terms to get our answer

this problem shows how complex

derivatives can be broken down into

simpler steps using the chain rule

9.3-1.10 Mastering the Chain Rule Differentiating (𝑠𝑖𝑛^3𝑥^3)  Step by Step

In this video, we break down the process of differentiating (𝑠𝑖𝑛^3𝑥^3)   using the chain rule. Learn how to handle composite functions and apply the power rule effectively. Perfect for students looking to grasp calculus concepts with clear, step-by-step explanations.
Hashtags:
#cpmrana  #Calculus #Differentiation #ChainRule #MathTutorial #Trigonometry


let's solve this complex differentiation

problem step by step we need to find the

derivative of s of 2 Aran of the sare <

TK of 1 - x over 1 + x instead of trying

to differentiate this directly which

would be quite complicated let's try to

simplify the expression first let's make

a substitution let x = cosine of 2 Theta

when we substitute this in our fraction

under the square root becomes 1 - cosine

2 thet over 1 plus cosine 2 Theta using

the double angle formula for cosine we

can write this in terms of s and cosine

Theta now under the square root we have

2 sin^2 thet over 2 cine 2 thet this

simplifies to tangent s Theta the square

root of tangent s thet is just tangent

thet now we have AR tangent of tangent

thet which just gives USA so 2 * AR

tangent of tangent Theta is just 2 Theta

therefore our expression simplifies to S

of 2 Theta remember that x equal cine of

2 Theta so we can substitute back to get

y = < TK of 1 - x^2 now we can

differentiate this much simpler

expression using the chain rule we get

1/ 2 * the square < TK of 1 - x^2 times

the derivative of what's inside the

square root the derivative of 1 - x^2 is

-2X simplifying this gives us our final

answer X over The Square < TK of 1 - x^2

and there we have it we've found the D

ative by first simplifying the original

expression using trigonometric

substitution and then differentiating

the simpler form

Understanding Trigonometric Differentiation [sin{2𝑡𝑎𝑛^(−1) √((1−𝑥)/(1+𝑥))}] A Step-by-Step Guide

Question: Differentiate the function  [sin{2𝑡𝑎𝑛^(−1) √((1−𝑥)/(1+𝑥))}]   with respect to x.
Dive into the world of trigonometric differentiation with this detailed video on differentiating a complex trigonometric function. We’ll break down each step, provide clear explanations, and ensure you have a firm grasp on how to handle functions like  [sin{2𝑡𝑎𝑛^(−1) √((1−𝑥)/(1+𝑥))}] . Perfect for students and math enthusiasts looking to master calculus concepts!

#Calculus #Trigonometry #Differentiation #MathTutorial #LearnMath


today we'll solve a challenging

derivative problem using logarithmic

differentiation let's find the

derivative of y = cent of X raised to

the tangent of X power the first step in

logarithmic differentiation is to take

the natural logarithm of both sides this

gives us natural log of yal natural log

of Cent x raed to the tangent X power

using the logarithm property that says

the natural log of a power equals the

power time the natural log of the base

we get natural log of Y = tangent x *

natural log of Cent X now we can

differentiate both sides with respect to

X the left side becomes DDX of natural

log Y and the right side becomes DDX of

tangent x * natural log of Cent X on the

left side we use the chain rule the

derivative of natural log Y is 1/ y *

dydx on the right side we use the

product rule since we're differentiating

a product let's evaluate the derivatives

on the right side the derivative of

natural log of Cent X is 1 / cang x *

cose 2 x and the derivative of tangent X

is secant 2 x

we can simplify 1 / Cent X to tangent X

this gives us tangent 2 x sin^2 X+

natural log Cent x * secant 2 X after

rearranging terms we get y * natural log

Cent x * secant 2 x - tangent 2 x/ sin^2

x using the identity that tangent 2 x =

sin^2 X over cine ^ 2 x we can simplify

this further to y * natural log Cent x *

SEC 2 x - secant 2 x finally

substituting back the original

expression for y we get our final answer

Cent x to the tangent X power * secant 2

x * the quantity natural log Cent X - 1

and that's how we use logarithmic

differentiation to find this challenging

derivative the key steps were taking the

natural log of both sides using the

product rule and carefully applying our

trigonometric identities

Solving Complex Derivatives: Finding d/dx of (cot x)^(tan x) Using Logarithmic Differentiation

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

In this comprehensive calculus tutorial, we break down the step-by-step process of finding the derivative of y = (cot x)^(tan x) using logarithmic differentiation. This video covers:
• Logarithmic differentiation technique
• Product and chain rules
• Trigonometric derivatives and identities
• Step-by-step solution walkthrough
• Common pitfalls and how to avoid them
Perfect for calculus students at university level or those preparing for advanced mathematics examinations. Watch how we transform a complex exponential function into a manageable form and solve it systematically.
#Calculus #Mathematics #Derivatives #LogarithmicDifferentiation #MathTutorial #CalcTutorial #MathEducation #TrigonometryCalculus


let's solve this complex differentiation

problem step by step we need to find the

derivative of e to the x * log base of X

first let's rewrite log base of X using

the change of Base formula log base of

xal log base e of x / log base e of a

which equals log Bas uh of e * log base

e of X now we can use the product rule

to differentiate this expression

remember the product rule states that

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

U * V Prime let's find the derivatives

of each part the derivative of e to X is

e to X and the derivative of natural log

of x is

1/x now let's apply the product rule we

multiply e to X by by 1 /x for the first

term and multiply natural log of x by E

to X for the second term then factor out

e to the X for our final step we

recognize that log base uh of e * log

base e of x equal log base a of X this

gives us our final answer e to x * thean

1 /x * log base uh of e plus log Bas of

X and that's how we differentiate e to x

* log Bas of X we use the change of Base

formula to rewrite the logarithm applied

the product rule and then simplified our

result using logarithm properties thanks

for watching

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

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

What You'll Learn:
- The Product Rule definition
- Step-by-step differentiation of e^x log_a(x)
- Breakdown of the formula and example problem-solving

This tutorial is perfect for calculus beginners or those revisiting 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 solve this calculus problem step

by step we need to differentiate the

logarithmic expression natural log of 1

+ < TK X over 1 - < TK x with respect to

X first let's use the logarithm property

that states the natural log of a

fraction equals the natural log of the

numerator minus the natural log of the

denominator this gives us the derivative

of natural log of 1 + < TK of x -

natural log of 1 - < TK X next we'll

apply the chain rule remember the

derivative of natural log of U with

respect to x = 1 / U * the derivative of

U with respect to X applying this to

both terms we get 1/ 1 + < TK x * the

Der of 1 + < TK x -1 / 1 - < TK X time

the derivative of 1 - < TK X now let's

find the derivatives of the inner

functions the derivative of square < TK

of X is 1 / 2 < TK X so 1 + < TK of X

differentiates to just 1 2 < TK X and 1

- < TK of X differentiates to 1/2 < TK X

let's simplify this expression the first

term becomes 1 / 2 < TK of x * 1 + < TK

X for the second term the negative signs

cancel out giving us + 1 / 2 < TK of x *

1 - < TK X to add these fractions we

need a common denominator when we

multiply both the numerator and

denominator by the respective missing

factors we get 1 - < TK of x + 1 + < TK

X all over 2 < TK x * 1 + < TK x * 1 - <

TK X for our final simplification notice

that 1 + < TK x * 1 - < TK of x = 1 - x

using the difference of squares formula

therefore our final answer is 1 / < TK x

* 1 - x

and there we have it we've successfully

differentiated this logarithmic

expression using properties of

logarithms the chain Rule and algebraic

simplification

Mastering Logarithmic Differentiation: Step-by-Step Explanation [𝒍𝒏((𝟏+√𝒙)/(𝟏−√𝒙))]

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

Dive into the world of logarithmic differentiation with our detailed step-by-step guide. Learn how to differentiate complex logarithmic functions using the chain rule and logarithm properties. Perfect for students looking to strengthen their calculus skills and understand logarithmic identities thoroughly.
#Calculus #Differentiation #MathTutorial #Logarithms #MathHelp


welcome today we'll solve a challenging

differentiation problem involving

exponential and inverse trigonometric

functions our task is to find the

derivative of e to x * sin inverse of X

let's break this down step by step step

one first we need to recognize that this

is a product of two functions e to X and

sin inverse of X this means we'll need

to use the product rule the product rule

states that the derivative of U * v = u

Prime * V + U * V Prime let's write this

out applying this to our problem we get

e to x * the derivative of sin inverse

of X Plus sin inverse of x * the

derivative of e to

X now we need to find each derivative

separately the derivative of e to the x

is simply e to X the derivative of sin

inverse of X is 1 over the < TK of 1 -

x^2 let's substitute these derivatives

into our equation we get e x * 1 < TK of

1 - x^2 plus sin inverse of x * e to

X finally we can write our answer as e x

over < TK of 1 - x^2 + e x * sin inverse

of X this is our final

answer let's verify why this makes sense

the first term comes from

differentiating e to the X and

multiplying by sin inverse of X using

the chain rule the second term comes

from keeping e to the x and

differentiating sign inverse of X both

terms contain e to the X because we're

differentiating a product involving e to

the X remember when differentiating

products of functions always one

identify the functions clearly two apply

the product rule correctly and three

know your basic derivatives especially

for exponential and inverse

trigonometric functions

thank you for watching

Understanding the Derivative of [𝑒^𝑥 𝑠𝑖𝑛^(−1)  𝑥]  Step-by-Step

Question: Differentiate [𝑒^𝑥 𝑠𝑖𝑛^(−1)  𝑥] with respect to x.
Dive into the step-by-step process of differentiating the function [𝑒^𝑥 𝑠𝑖𝑛^(−1)  𝑥] . Explore how to apply the product rule and differentiate inverse trigonometric and exponential functions with clear explanations.
#Calculus #Differentiation #MathTutorial #ProductRule #InverseTrigFunctions #ExponentialFunctions


let's solve this complex differentiation

problem step by step we'll start with

the function y = sin s of the natural

log of cine X let's break this down into

manageable Parts first notice that this

is a composite function with multiple

layers we have sin squ as the outer

function natural log as the middle

function and cine X as the innermost

function to differentiate this we'll use

the chain rule multiple times let's

start with the outer function the

derivative of sin squ is 2 * s time the

derivative of s next we apply the chain

rule again to differentiate s of natural

log of cosine X this gives us cosine of

natural log of cosine x times the

derivative of the natural log of cosine

X here's where we can use a helpful

trigonometric identity not that 2 * sin

* cosine equals s of 2 * the angle this

lets us simplify the first part of our

expression now we need to differentiate

the natural log of cosine X remember the

derivative of natural log is 1 over the

function time the derivative of the

function the derivative of cosine X is

sinx we can rearrange this to get sin 2

* natural log of Co

x * sin x cosine X finally we recognize

that sin x cosine X is the tangent

function giving us our final

answer tangent x * sin 2 natural log of

cosine X this problem showcases the

power of the chain Rule and how we can

use trigonometric identities to simplify

complex expressions

Differentiating Sin Squared of Log Cosine [𝑠𝑖𝑛^2 {ln(cos𝑥)}] Step-by-Step Guide

Join us as we delve into differentiating  [𝑠𝑖𝑛^2 {ln(cos𝑥)}]  using essential calculus techniques. This video provides a clear, step-by-step explanation to help you master derivatives involving trigonometric and logarithmic functions, perfect for students and educators seeking in-depth comprehension.
#Calculus #Derivative #Trigonometry #MathTutorial #ChainRule #LogarithmicFunctions #STEM #AdvancedMath #MathHelp #Differentiation


welcome today we'll solve a challenging

derivative problem using logarithmic

differentiation let's find the

derivative of y = x raised to the

inverse tangent of X this is a complex

function because we have a variable base

X and a variable exponent involving

inverse tangent the best strategy here

is to use logarithmic differentiation

we'll take the natural log of both sides

and then differenti iate let's solve the

step by step First We Take the natural

log of both sides using the properties

of logarithms we can write this as

inverse tangent of x * natural log of x

now we can differentiate both sides with

respect to X on the left side we use the

chain rule the derivative of natural log

of Y is 1 / y * Dy DX on the right side

we use the product rule since we're

differentiating the product of two

functions we get inverse tangent of x *

the derivative of natural log of x plus

natural log of x times the derivative of

inverse tangent of X now let's

substitute the known derivatives the

derivative of natural log of x is 1 /x

and the derivative of inverse tangent of

X is 1 1 + x^2 to solve for dydx we

multiply both sides by y remember that y

= x to the inverse tangent of x from our

original equation and there we have it

the derivative of x to the inverse

tangent of x = x to the inverse tangent

of x * the sum of inverse tangent of x /

x + natural log of x X over 1 + x^2 this

problem shows the power of logarithmic

differentiation in handling complex

expressions with variable bases and

exponents remember whenever you see a

function with a variable base and

variable exponent logarithmic

differentiation is often the best

approach

Mastering Logarithmic Differentiation: Solving [y = x^(arctan x)]

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

Learn how to find the derivative of y = x^(arctan x) using logarithmic differentiation in this comprehensive math tutorial. We break down this complex differentiation problem into clear, easy-to-follow steps using the natural logarithm and chain rule. Perfect for calculus students and math enthusiasts!
Key topics covered:
• Logarithmic differentiation technique
• Properties of natural logarithm
• Chain rule application
• Product rule
• Derivatives of arctan(x) and ln(x)
• Step-by-step solution process
#Mathematics #Calculus #LogarithmicDifferentiation #MathTutorial #ChainRule #ProductRule #Derivatives #MathEducation #MathHelp #CalcTutorial


let's find the derivative of cosine 3x

using the first principles of calculus

we'll go through this step by step let's

start by defining our function let F ofx

= cine 3x therefore f of x + H = cosine

3 * x + H by definition the derivative

is the limit as H approaches Z of f ofx

+ H - FX all divid H substituting our

function we get the limit as H

approaches Z of cosine 3 * the quantity

x + H - cine

3x all divid H now we'll use the

trigonometric identity for the

difference of cosiness cosine of minus

cosine B = -2 * sin of a plus b/ 2 * sin

of A- B 2 applying this identity to our

expression we get -2 * sin of 3x + 3 H +

3x / 2 * sin of 3x - 3x - 3 h / 2 all ID

by H simplifying the Expressions inside

the S functions we get 2 * sin of 6X +

3H / 2 * sin of -3 h / 2 all / H further

simplifying we get 2 * sin of 3x + 3 h /

2 * sin of 3H / 2 all divided H now we

can separate this into two limits the

limit as H approaches Z of s of 3x + 3 h

/ 2 times the limit as H approaches 0 of

netive s of 3 h / 2 / 3 h / 2 * 3 as h

approaches zero the first limit becomes

sin 3x and the second limit equals -1

due to the fundamental limit of sin

Theta over Theta as Theta approaches

zero therefore the derivative of cosine

3x = -3 sin 3x this result matches the

chain Rule and the standard derivative

of cosin x which is sinx the factor of 3

comes from the chain rule when

differentiating the inside function 3x

Understanding the Derivative of cos 3x  Using First Principles

Dive into the fundamentals of calculus as we explore how to determine the derivative of the exponential function sin3x 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're going to solve a

challenging differentiation problem step

by step our task is to differentiate y =

theare < TK of 1 - x over 1 + x let's

break this down step by step first we

can rewrite this as the derivative with

respect to X then we can rewrite this as

a quotient of square roots to make it

easier to handle now comes the key part

we need to use the quotient rule here

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 U is the square < TK of 1 - x

and V is the square < TK of 1 + x this

gives us this longer

expression now we need to find the

derivatives of our square root terms

remember the chain rule for square roots

the derivative of square root of x is 1

/ 2 < TK x times the derivative of

what's inside so the derivative of

square T of 1 - x is - 1 /2 < TK of 1 -

x and the derivative of sare < TK of 1 +

x is 1 / 2 < TK of 1 + x let's simplify

this expression first we can factor out

-1 over 1 + x then we can combine the

terms inside the brackets by finding a

common denominator the numerator

simplifies to just two and we can write

the denominator more compactly our final

answer is -1 over the 1 + x * the < TK

of 1 - x^2

let's review the key steps we used first

we rewrote the expression as a quotient

of square roots then we applied the

quotient rule next we used the chain

rule to find the derivatives of the

square roots finally we simplifi to get

our answer

How to Differentiate a Square Root Quotient Step-by-Step| [√((1−𝑥)/(1+𝑥))]

Question
Differentiate the expression: [√((1−𝑥)/(1+𝑥))]

Unlock the secrets to differentiating complex expressions with our in-depth, step-by-step breakdown of the derivative of a square root quotient. Learn how to apply the quotient rule and simplify expressions effectively. Perfect for students and math enthusiasts looking to master calculus!
#Calculus #Differentiation #MathTutorial #LearnMath #QuotientRule


let's solve this differentiation problem

step by step using the chain rule we

need to find the derivative of the

function 1 + sin of 2x all 2ar to solve

this using the chain rule let's break it

down into steps first let's make a

substitution let U equal 1 + sin of 2x

then our function can be Rewritten as y

= u^2 now according to the chain rule

the derivative of y with respect to x

equals the derivative of y with respect

to U times the derivative of you with

respect to X let's find each part

separately first the derivative of y

with respect to U since y = u ^2 this

derivative is 2 U we'll keep this part

for now and find the second part next

let's find the derivative of you with

respect to X remember that u = 1 + S of

2x the derivative of 1 is zero and the

derivative of sin of 2x is 2 cosine of

2x using the chain rule

again now let's multiply these parts

together according to the chain rule we

have 2 * 2 cosine of 2x remember that U

is 1 + s of 2x so let's substitute that

back

in and there we have our final answer

the derivative of 1 + sin of 2x^2 is 4 *

1 + sin of 2x * cine of 2x let's verify

our

steps the key to solving this problem

was breaking it down into simpler Parts

using the chain rule

and then carefully putting those parts

back together remember when dealing with

composite functions like this the chain

rule is your best friend it allows us to

break down complex derivatives into

simpler manageable steps

Mastering Calculus Differentiating {(1+sin2𝑥)^2}  Step by Step

In this video, we delve into the process of differentiating the function {(1+sin2𝑥)^2}  with respect to . We’ll guide you through each step using the chain rule and fundamental differentiation techniques. Whether you’re a student brushing up on calculus or keen on mastering differentiation, this detailed breakdown will enhance your mathematical understanding.
#Calculus #Differentiation #ChainRule #MathTutorial #Mathematics #Educational


let's solve this calculus problem step

by step we need to differentiate the

inverse cosine of 1 - 2x with respect to

X to solve this we'll use the chain Rule

and the derivative formula for inverse

cosine remember the derivative of

inverse cosine of U is- 1 over theare <

TK of 1 - U ^2 * the derivative of U in

our case u = 1 - 2 X let's substitute

this into our formula now let's find the

derivative of U with respect to X the

derivative of 1 - 2x is -2 next let's

expand what's inside the square root

under the square root we have 1us the

quantity 1 -

2x^2 when we expand this we get 1us the

quantity 1 - 4x +

4x^2 let's simplify what's under the

square root 1 - 1 + 4x - 4x^2 simplifies

to 4x -

4x^2 now we can factor out a 4 from

under the square root the square < TK of

4x - 4x^2 = 2 * the < TK of x - x^2

finally we can cancel the twos in the

numerator and denominator this gives us

our final answer 1/ < TK x - x^2 to

verify this is correct we can check that

we used the chain rule correctly

expanded the squared term properly

simplifi the expression under the square

root and cancel common factors each step

followed logically from the previous one

leading us to our final answer

Mastering Derivatives: Simplifying the Inverse Cosine Function [𝑐𝑜𝑠^(−1) (1−2𝑥)]

Join us as we explore the differentiation of the inverse cosine function, [𝑐𝑜𝑠^(−1) (1−2𝑥)]. Follow along with our detailed, step-by-step breakdown to understand each part of the process and learn how to simplify derivatives effectively. Perfect for math students aiming to solidify their calculus skills!
#Calculus
#Derivatives
#MathTutorial
#LearningMath
#Trigonometry


let's solve this calculus problem step

by step we need to find the derivative

of the square root of tangent Theta

first let's rewrite this in a form

that's easier to work with the square

root is equivalent to raising something

to the 1/2 power so square root of

tangent Theta can be written as tangent

Theta to the 1/2 power now we can use

the power rule for derivatives the power

rule rule states that when

differentiating x to the N power we

multiply by the power and reduce the

exponent by one however since we're

differentiating with respect to Theta

and have a tangent function inside we'll

also need to use the chain rule let's

apply the power rule first we multiply

by 1/2 and reduce the power by 1 this

gives us 12 * Tang thet to the2 power

now for the chain rule part we need to

multiply by the derivative of the inside

function which is tangent Theta the

derivative of tangent Theta is secant s

thet putting it all together we get 12 *

Tang thet to -2 power * SEC 2 thet this

can be Rewritten as 1 / 2 < TK of

tangent thet * secant 2 thet which is

our our final answer to review we solve

this by first rewriting the square root

as a power then using the power rule and

chain rule together the key steps were

reducing the power by one multiplying by

the power and including the derivative

of the inside function

9.3-1.6 How to Differentiate (√tan𝜃)   Step by Step Explanation.

In this video, we break down the process of differentiating the function (√tan𝜃)  using the chain rule and power rule. Follow along as we explain each step and simplify the expression to find the derivative. Perfect for students and math enthusiasts looking to improve their calculus skills!
#Calculus #Differentiation #MathTutorial #ChainRule #PowerRule #Trigonometry #cpmrana


welcome today we'll learn how to find

the derivative of a complex function

using logarithmic differentiation this

technique is particularly useful when

dealing with products quotients and

exponents let's look at our question we

need to find the derivative of y with

respect to X where y equals this complex

fraction the first step in logarithmic

differentiation is to take the natural

log of both sides this gives us Ln y

equal Ln of our entire fraction next we

can use the properties of logarithms to

break this down the log of a quotient is

the difference of logs so we get Ln of

the numerator minus Ln of the

denominator now we can further break

this down using more log properties the

log of a product is the sum of logs and

the log of a power is the power * the

log we get 2 Ln of x + 1

+ Ln of x - 1 - 3 Ln of x + 4 - x Ln of

e now comes the key step we

differentiate both sides with respect to

X remember that when we differentiate Ln

y we get 1 / y *

dydx let's differentiate the right side

term term by term remember that the

derivative of lnx is 1 /x after applying

the chain rule we get 2x + 1 + 1 / 2 *

1/x - 1 - 3/ x + 4 -1 now we can solve

for

dydx multiply both sides by Y and we get

our derivative equals y * this

expression in Brackets

finally we substitute back the original

expression for y this gives us our

complete derivative while this looks

complicated we found it using a

systematic approach that broke down a

complex problem into manageable steps

and that's how we use logarithmic

differentiation to find the derivative

of a complex function thank you for

watching

Mastering Logarithmic Differentiation [((𝑥+1)^2  √(𝑥−1))/((𝑥+4)^3  𝑒^𝑥 )] Step-by-Step Solution

Question
Given the function y= [((𝑥+1)^2  √(𝑥−1))/((𝑥+4)^3  𝑒^𝑥 )]  , find dy/dx using logarithmic differentiation.

Unlock the secrets of logarithmic differentiation with this detailed, step-by-step tutorial. We break down the process of finding the derivative of complex functions using logarithms, making each step clear and easy to follow. Perfect for students and math enthusiasts looking to enhance their calculus skills.
#Calculus #LogarithmicDifferentiation #MathTutorial #Derivatives #MathHelp


let's find the derivative of tangent 3x

using the first principle's definition

of derivatives we'll start by defining

our function f ofx = tangent 3x the

definition of a derivative states that F

Prime of x equals the limit as H

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

over H when we substitute our function

we get the limit as H approaches is 0 of

tangent of 3 * x + H - tangent of 3x all

over H next we'll convert the tangent

functions to S over cosine this gives us

the limit as H approaches 0 of sin of 3x

+ H over cosine of 3x +

hus s of 3x over cosine of 3x all over H

to subtract these fractions we need a

common denominator

multiplying the numerator and

denominator of each fraction by the

denominator of the other fraction gives

us this expression the numerator is in

the form of the sign difference formula

this simplifies to S of 3 H in the

numerator now we can factor out a three

from the S term giving us 3 * s of 3 H

over 3 H ided by the product of the

cosiness as H approaches 0

s of 3 H over 3 H approaches 1 and

cosine of 3x + H approaches cosine of 3x

this gives us 3 over cosine 2 3x which

equal 3 secant 2 of 3x therefore we have

shown that the derivative of tangent of

3x = 3 secant 3x

Understanding the Derivative of tan3x  Using First Principles

Learn the basics of calculus by finding the derivative of the exponential function tan3x using first principles. Follow our step-by-step explanations and examples to deepen your understanding of derivatives and exponential functions.
#Calculus #Derivatives #ExponentialFunctions #MathTutorial #FirstPrinciples #tan3x


let's learn how to differentiate cosine

of pix over

180 this is an important expression that

helps us convert between degrees and

radians let's start with our expression

we have y = cosine of x de which we can

write as y = cosine of pkx over

180 to differentiate this we'll need to

use the chain rule the chain rule states

that if yal F of G ofx

then Dy DX = frime of G ofx * G Prime of

X let's identify our functions in our

case the outer function f of U is cosine

of U and the inner function G ofx is pix

over

180 now let's apply the chain rule step

by step first we need fime of G ofx the

derivative of cosine is negative s so

this gives us negative sign of PX over

180 next we need G Prime of X which is

the derivative of pix over

180 this gives us pi over

180 finally we multiply these together

according to the chain rule this gives

us our final answer Dy DX equal negative

s of PX over 180 * < over 18 80 let's

visualize what this means here's the

graph of y equal cine of PX over 180 in

blue and its derivative < over 180 * s

of pix over 180 in red notice how the

derivative graph shows the rate of

change of the original function when

cosine is increasing its derivative is

positive and when cosine is decreasing

its derivative is negative the

derivative is zero exactly at the Peaks

and troughs of the cosine function and

that's how we differentiate cosine of PX

over 180 using the chain rule thank you

for watching

9.3-1.8 Understanding Trigonometric Differentiation (𝑐𝑜𝑠𝑥^0) Step-by-Step Guide Using the Chain Rule

In this video, we’ll explore how to differentiate trigonometric functions using the chain rule. Learn to tackle expressions like (𝑐𝑜𝑠𝑥^0) with ease. We’ll break down the steps, helping you understand the transformation of degrees to radians, applying the chain rule, and combining derivatives efficiently. Perfect for students and math enthusiasts looking to strengthen their calculus skills!
Hashtags:
#cpmrana  #Calculus #Trigonometry #ChainRule #MathTutorial #Differentiation


welcome today we're going to learn about

logarithmic differentiation by finding

the derivative of a complex exponential

function our function is y = x ra to the

power of E to X why do we need

logarithmic differentiation here well

this function is particularly tricky

because we have a variable X as both the

base and in the exponent using normal

differentiation rules would be extremely

complicated however by taking the

natural log of both sides first we can

make this much easier to handle let's

solve a step by step step one take the

natural log of both sides this gives us

Ln of Y = Ln of x raed to e to X Step 2

use the logarithm property that says Ln

of x raed to n = n * Ln of X this gives

us Ln of y = e x * Ln of

X step three 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 step four for the right side we

need to use the product rule since we

have e to x * Ln of X the product rule

says that the derivative of U * v = u *

dvdx + B * D DX in our case U is e to

the X and V is Ln of

X step five let's differentiate each

part the derivative of e to X is e to X

the derivative of Ln of X is 1 /x

putting this together we get e to x * 1

/x + Ln of x * e to

X step six now we can solve for

dydx multiply both sides by 1 Y which is

X the E to X this gives us our final

answer Dy DX = x e to the x * the

quantity e to X over x + Ln of x * e to

X and there you have it we've found the

derivative using logarithmic

differentiation this method is is

particularly useful when dealing with

functions that have variables in both

the base and exponent thank you for

watching

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

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

📚 In this comprehensive calculus tutorial, we break down how to find the derivative of y = [〖𝒙^𝒆〗^𝒙] using logarithmic differentiation. Perfect for calculus students and math enthusiasts!
🎯 What You’ll Learn:
• Why logarithmic differentiation is crucial for complex exponential functions
• How to properly apply natural logarithm properties
• Step-by-step application of chain rule and product rule
• Clear explanation of each mathematical move
• Common pitfalls to avoid
💡 Related topics:
Chain Rule
Product Rule
Natural Logarithms
Exponential Functions
Calculus I
📝 Need more practice? Check out our full calculus playlist!
#Calculus #Mathematics #LogarithmicDifferentiation #ChainRule #MathTutorial #CalcTutorial #MathHelp #ExponentialFunctions #CalculusHelp #Education


let's solve this challenging derivative

problem step by step we need to find the

derivative of Ln of sinx s raised to the

power n this is a complex function

involving multiple layers first we have

x s inside s then sin x^2 is inside a

natural log and finally all of that is

raised to the power n we'll need to use

the chain rule multiple times let's

start with the outermost layer using the

chain rule we bring down the power n and

reduce the exponent by one this gives us

n * Ln of sin x^ 2^

nus1 time the derivative of what's

inside now we need to find the

derivative of Ln of sin x^2 using the

chain rule again this is 1/ sin x^2 *

the derivative of sin x^2 let's handle

the derivative of sin

x^2 using the chain rule one more time

this is cosine x^2 time the derivative

of X2 the derivative of x^2 is simply 2x

let's substitute that in now we can

simplify this expression notice that

cine x^2 over sin x^2 is the definition

of cotangent x^2 rearranging the terms

gives us our final answer 2 NX * cang

x^2 * Ln of sin x^ 2^ n -1 let's review

the key steps we used first we used the

chain rule for the power function then

we used it again for the natural log

finally we used it one more time for the

sign function this problem shows how the

chain rule can help us tackle complex

derivatives by breaking them down into

simpler steps

Mastering the Derivative of Logarithmic Trigonometric Functions [ln(sin𝑥^2)]^𝑛

Unlock the secrets of differentiating logarithmic trigonometric functions with this step-by-step tutorial! Learn how to find the derivative of expressions like [ln(sin𝑥^2)]^𝑛 using product and chain rules. This video is perfect for students tackling calculus problems or anyone looking to deepen their understanding of advanced mathematical concepts.
#Calculus #Differentiation #Trigonometry #MathTutorial #AdvancedMath


let's solve this challenging derivative

problem step by step we'll use the chain

Rule and inverse function rules to find

the derivative of tan inverse of sin of

e to X let's break this down into parts

we have three nested functions first the

outer function is tan inverse then we

have S as our middle function and

finally e to thex as our innermost

function now let's recall the derivative

formula for in

tangent the derivative of tan inverse of

U with respect to U is 1/ 1 + U ^2 in

our case U equal s of e to X so when we

apply the inverse tangent rule we get 1

1 + sin^2 of e to X time the derivative

of sin of e to X now we need to find the

derivative of sign of e to X using the

chain rule the D dtive of s of V is

cosine of V * the derivative of v in our

case v is e to X so we get cosine of e

to the X time the derivative of e to the

X finally we know that the derivative of

e to X is just e to X so our final

answer is e to x * cosine of e to X all

over 1 + sin 2 e to X let's let's verify

that this makes sense our derivative is

a fraction where the numerator contains

e to X which came from the derivative of

our innermost function multiplied by

cosine of e to X which came from the

derivative of s the denominator comes

from the derivative formula of inverse

tangent this is exactly what we'd expect

from applying the chain rule to our

nested functions

Mastering Derivatives: Calculating the Derivative of [𝑡𝑎𝑛^(−1) (𝑠𝑖𝑛𝑒^𝑥 )]

In this video, we break down the step-by-step process of finding the derivative of [𝑡𝑎𝑛^(−1) (𝑠𝑖𝑛𝑒^𝑥 )] . Learn how to apply the chain rule and understand the logic behind each step. Perfect for calculus students looking to strengthen their derivative skills!
#Calculus #Derivatives #MathTutorial #ChainRule #MathHelp #InverseTrigonometricFunctions


let's solve this challenging

differentiation problem step by step we

need to find the derivative of sign of

natural log of secant x with respect to

X let's break this down using the chain

rule when we have a composite function

like this we multiply the derivative of

the outer function by the derivative of

the inner function here s is our outer

function and natural log of secant x is

our inner function for our first step we

know that the derivative of s is cosine

so we'll have cosine of the inner

function times the derivative of the

inner function now we need to find the

derivative of natural log of secant x

using the chain rule again we know that

the derivative of natural log of U is 1/

U times the derivative of U the final

piece we need is the derivative of

secant x which is secant x * tangent X

now let's simplify the secant x terms

cancel in the middle leaving us with

cosine of natural log of secant x *

tangent X therefore our final answer is

tangent x * cine of natural log of

secant x this problem demonstrated the

power of the chain Rule and how breaking

down complex derivatives into smaller

steps makes them manageable remember to

identify your composite functions and

apply the chain rule systematically

Differentiating [sin𝑙𝑛(sec𝑥 ) ] - Step-by-Step Calculus Explained!

In this video, we break down how to differentiate the expression [sin𝑙𝑛(sec𝑥 ) ]  step by step. Using the chain rule and understanding derivatives, we simplify the process to make it easier to understand. Perfect for students seeking a deeper comprehension of calculus concepts!
#Calculus
#MathTutorial
#Differentiation
#ChainRule
#MathHelp


today we'll find the derivative of

secant 2x using the first principles

approach let's break this down step by

step let's start by defining our

function let F ofx = secant 2x then f of

x + H = secant of 2 * x + H by

definition the derivative is the limit

as H approaches Z of f of x + h -

FX all divid H substituting our function

we get the limit as H approaches 0 of

secant of 2 * x + H - secant of 2x all

ID H since cant is 1 / cosine we can

rewrite this as the limit as H

approaches Z of 1/ cosine of 2x + H - 1

/ cosine of 2x all divided by H to

subtract these fractions we need a

common denominator multiplying both

terms by their respective denominators

gives us cosine of 2x - cosine of 2x + H

over the product of the cosiness now we

can use the cosine difference formula

the difference of cosiness equals -2 * s

of their sum / 2 * s of their difference

/ 2 simplifying the arguments in the

sign terms we get S of 2x + H * sin of H

in the numerator we can split this into

separate limits the limit of s of 2x + H

as H approaches Z is s of 2x the limit

of sign H over H as H approaches Z is 1

and the limit of cosine of 2x + H is

cosine of 2x finally we can write this

as 2 sin of 2x/ cosine ^ 2 2x which = 2

tangent of 2x * secant of 2x therefore

the derivative of secant of 2x = 2

tangent of 2x * secant of 2x and that's

how we find the derivative of secant 2x

using first principles thank you for

watching

Understanding the Derivative of sec2x  Using First Principles

Master calculus fundamentals by deriving the exponential function sec2x from first principles. Our step-by-step explanations and examples will strengthen your understanding of derivatives and exponential functions.
#Calculus #Derivatives #ExponentialFunctions #MathTutorial #FirstPrinciples #sec2x


let's solve this calculus problem step

by step we need to find the derivative

of y = 10 x - Cent X over Tan x + Cent X

the first step is to convert tangent and

coent into their s and cosine forms

tangent X is sin x over cosine x and

cang x is cine X over sin x next we'll

find a common denominator in both the

numerator and denominator in the

numerator we multiply both fractions by

sin x cine X this gives us sin^2 x -

cosine 2 x/ sin x cosine X similarly in

the denominator we get sin^2 x + cine 2

x/ sin x cine X now let's recall a

fundamental trigonometric identity sin^2

X Plus cosine 2 x = 1 this means our

denominator simplifies to 1 this

simplifies to Just s^ 2 x - cosine 2 x

and here's another useful identity sin 2

x - cosine ^ 2 x = cosine of 2x now that

we have y = cosine of 2x we can find its

derivative the derivative of cosine is

negative sign and when we have a chain

rule situation with 2x we multiply by

the derivative of the inside function

which is two finally we can simplify

this the two negatives cancel out and we

get our final answer the derivative

equals 2 sin 2x let's verify this makes

sense our original function was a ratio

of trigonometric functions that

simplified to cosine 2X taking the

derivative gave us 2 sin 2x which

matches what we'd expect from the

derivative of a cosine function with a

factor of 2x inside

Mastering Trigonometric Differentiation {(tan𝑥 −cot𝑥)/(tan𝑥+cot𝑥)} Step-by-Step Guide

Unlock the secrets of trigonometric differentiation with our comprehensive tutorial! In this video, we will walk you through the process of differentiating the expression y={(tan𝑥 −cot𝑥)/(tan𝑥+cot𝑥)}   step-by-step. Whether you’re a math enthusiast or preparing for exams, this guide will help you understand each step with clear explanations. Subscribe for more math tutorials and enhance your problem-solving skills.
#Trigonometry #Calculus #MathTutorial #Differentiation #MathHelp #StudyMath #EducationalVideo #MathSkills


let's solve this differentiation problem

using the quotient rule we'll go through

this step by step let's start with our

function we need to differentiate F ofx

= 3x + 5 / 2x^2 - 9 to do this we'll use

the quotient rule the quotient rule

states that the derivative of U over V

equals V * the derivative of U minus U *

the derivative of V all over v²

let's identify our u and v here U = 3x +

5 and V = 2x^2 - 9 now let's find the

derivatives of u and v the derivative of

U is 3 and the derivative of V is 4X

let's substitute these into our quotient

rule formula we get 2x^2 - 9 * 3 - 3x +

5 * 4X all over 2x^2 - 9^ 2 now let's

expand the numerator first 2x^2 - 9 * 3

gives us 6 x^2 - 27 then 3x + 5 * 4X

gives us 12 x^2 + 20x so in the

numerator we have 6 x^2 - 27 - 12x^2 -

20x this simplifies to -6 x^2 - 20x - 27

therefore our final answer is - 6x^2 +

20x + 27 over 4X 4 - 36 x^2 + 81 we can

factor out a negative from the numerator

to write it as * the 6 x^2 + 20x + 20 7

over 4X 4 - 36 x^2 + 81 and there we

have it we've successfully used the

quotient rule to find the derivative of

this function remember the key steps

identify u and v find their derivatives

apply the quotient rule formula and then

simplify the result

How to Differentiate Rational Functions ((𝟑𝒙+𝟓)/(𝟐𝒙^2-9)) Step-by-Step Tutorial

In this video, we dive into the differentiation of rational functions using the quotient rule. Follow along as we break down each step to find the derivative of ((𝟑𝒙+𝟓)/(𝟐𝒙^2-9)). Master calculus with our clear and concise explanations, and boost your math skills today!
#Calculus #Differentiation #MathTutorial #QuotientRule #LearnMath


welcome today we'll solve a challenging

differentiation problem using

logarithmic

differentiation our task is to find the

derivative of x to the X Power Plus sin

x raised to the natural log of x let's

break this into two parts to make it

easier we'll call the first part y1 = x

x and the second part Y 2 = sinx to the

natural log of x let let's start with y1

to differentiate this we'll use

logarithmic

differentiation first take the natural

log of both sides using the properties

of

logarithms this simplifies to x * lnx

now we can differentiate both sides with

respect to X on the left side we get 1

over y1 * dy1 DX on the right side using

the product rule we get x * d by DX of

Ln x + Ln x * d by DX of X simplifying

the right side we get 1 + Ln X then

multiply both sides by y1 to get Dy 1 DX

= y1 * 1 + lnx finally substitute back

y1 = X to X now let's tackle Y 2 again

we'll take the natural log of both sides

this gives us Ln Y2 = Ln x * Ln of sinx

differentiating both sides we get 1 / Y2

* dy2 DX = Ln x * the derivative of Ln

sin x + Ln sin x * the derivative of Ln

X after simplifying we get Ln X Cent X

Plus Ln sin x

/x now we can combine our results the

derivative of the original expression is

the sum of Dy 1 DX and dy2 DX this gives

us x x * 1 + Ln X Plus sin x to the Ln x

* Ln xent x + Ln sin x /x and that's our

final answer we used logarithmic

differentiation to handle these complex

expressions with variables in the

exponents this technique is very

powerful for differentiating such

expressions

Mastering Complex Derivatives: How to Find d/dx[x^x + (sin x)^ln x] | Step by Step

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

📚 In this detailed calculus tutorial, we break down the solution to finding the derivative of x^x + (sin x)^ln x using logarithmic differentiation. Perfect for advanced calculus students and those preparing for higher mathematics.
Key topics covered:
• Logarithmic differentiation technique
• Chain rule and product rule applications
• Breaking complex derivatives into manageable parts
• Step-by-step solution methodology
#Calculus #Mathematics #DerivativeCalculus #MathTutorial #CalcTutorial #MathEducation #AdvancedMath


welcome today we'll solve a challenging

derivative problem involving inverse

trigonometric functions let's look at

our question this is a complex

expression involving inverse tangent to

solve it we'll use a clever substitution

technique let's substitute xal secant T

this means tal inverse secant x now

let's substitute this into our original

expression inside the square root

x^2 - 1 becomes SEC 2 T -1 this is equal

to tangent 2 T which is a fundamental

trigonometric identity this simplifies

to 1 / tangent T which is coent T and

coent T can be written as tangent of Pi

/ 2us T here's where things get

interesting the inverse tangent and

tangent cancel each other out leaving

with just piun / 2 - t now we can

differentiate with respect to X the

derivative of Pi / 2 is 0 and the

derivative of inverse secant x is 1 /x *

the < TK of x^2 -1 therefore our final

answer is -1 /x * the < TK of x^2 - 1

this problem showcases the power of

clever sub subtitution and trigonometric

identities in calculus the key was

recognizing that we could simplify our

expression using the cant substitution

Understanding the Derivative of Inverse Trigonometric Functions [𝑡𝑎𝑛^(−1)  1/√(𝑥^2−1)]

Question
Find the derivative of the function [𝑡𝑎𝑛^(−1)  1/√(𝑥^2−1)] .

In this video, we dive deep into the derivative of inverse trigonometric functions, focusing on how to differentiate  [𝑡𝑎𝑛^(−1)  1/√(𝑥^2−1)]  . We’ll explore the process step-by-step, using trigonometric identities and calculus techniques to derive the final result. Perfect for students and enthusiasts looking to strengthen their calculus skills!
#Calculus #Trigonometry #Derivatives #MathTutorial #InverseTrigFunctions


welcome today we'll solve a challenging

derivative problem step by step our task

is to find the derivative of the natural

logarithm of x- the < TK of x^2 - 1

let's break this down step by step first

we'll use the chain rule since we have a

natural logarithm of a complex

expression remember the derivative of

natural log is one over the function

times the derivative of the inner

function now we can differentiate the

inner expression the derivative of x is

1 and for the square root term we'll

need to use the chain rule again for the

square root term we use the power rule

the derivative of square < TK of U is 1

/ 2 < TK U * the derivative of U the

derivative of x^2 - 1 is 2x let's

substitute that in now we can simplify

the expression inside the brackets the

two and two cancel out giving us this

simpler form to combine these terms we

need a common denominator we can rewrite

one as the square root of x^2 - 1/

itself next we can factor out -1 from

the numerator and finally notice that x

- < TK of x^2 - 1 appears in both both

numerator and denominator so they cancel

out leaving us with our final answer -1

over < TK of

x^2-1 this was a complex derivative

problem that required careful

application of multiple derivative rules

the chain rule the power rule and

careful algebraic manipulation the key

was breaking it down into manageable

steps and working through it

systematically

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

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


let's solve a complex derivative problem

step by step we need to find the

derivative of y = e power of x^2 + x^ of

x^2 this is a complex function that will

require multiple differentiation

techniques let's break it down into

simpler Parts by substituting U equal e

to the power of x^2 and v = x to the

power of x^2 let's start with finding

dudx we'll use logarithmic

differentiation taking the natural log

of both sides Ln of U = Ln of e to power

of x^2 this simplifies to x^2 since Ln

of eal 1 now we can differentiate both

sides with respect to X using implicit

differentiation we get 1 / U * D DX = 2X

solving for DX we get DX = U * 2x

substituting back u = e power of x^2 we

get Dux = 2x * E power of x^2 now for

the second part finding

dvdx again we'll use logarithmic

differentiation taking the natural log

of both sides Ln of V = Ln of x to the

power of x^2 = x^2 Ln X differentiating

both sides implicitly we get 1 / V *

dvdx = x^2 * 1 /x + Ln x * 2x solving

for dvdx we get dvdx = V * the x + 2x

lnx substituting back v = x to^ of x^2

and simplifying we get dvdx = X the

power of x^2 + 1 * the quantity 1 + 2 Ln

X finally we can combine our results

using the sum rule of differentiation

the derivative D ydx = DX + dvdx which

gives us 2x * e^ of x^2 + x ^ of x^2 + 1

* the quantity 1 + 2 lnx this problem

demonstrated several key calculus

Concepts logarithmic differentiation

implicit differentiation the chain Rule

and the sum rule of derivatives by

breaking down the complex function into

simpler parts and using these techniques

systematically we were able to find the

derivative of this challenging

function e

Advanced Calculus: Finding the Derivative of e^(x²) + x^(x²) | Step-by-Step Solution

Question:
Find the derivative of y = e^(x²) + x^(x²)

Learn how to find the derivative of a complex function combining exponential and power functions. In this detailed tutorial, we’ll break down the solution of dy/dx where y = e^(x²) + x^(x²) using:
• Logarithmic differentiation
• Chain rule
• Product rule
• Implicit differentiation
• Step-by-step explanations
• Clear mathematical reasoning
Perfect for:
📚 Calculus students
🎓 University mathematics
📐 Advanced differentiation techniques
💡 Math problem-solving strategies
#Calculus #Mathematics #Derivatives #MathTutorial #ComplexDerivatives #CalulusProblems #MathEducation


let's learn how to find the derivative

of a quotient using the quotient rule

we'll find the derivative of f ofx =

3x^2 + 5 / 2x + 1 first let's recall the

quotient rule formula the derivative of

U over v = v * the derivative of uus U *

the derivative of V all over V ^2 let's

identify u and v in our function here U

of X is 3x^2 + 5 and V of X is 2x + 1

next we need to find the derivatives of

u and v the derivative of U is 6X and

the derivative of V is 2 now let's

substitute everything into the quotient

rule formula we'll have 2X + 1 * 6x -

3x^ 2 + 5 * 2 all over 2x + 1 2

let's simplify the numerator step by

step first let's multiply 2x + 1 by 6X

that gives us 12 x^2 + 6 x next let's

multiply 3x^2 + 5x 2 that gives us 6 x^2

+ 10 now let's substitute these back

into our derivative expression we have

12 x^2 + 6 x - 6 x^2 - 10 all over 2x +

1^ 2 let's simplify the numerator 12 x^2

- 6 x^2 gives us 6 x^2 we keep the plus

6X and -10 therefore our final answer is

6 x^2 + 6 x - 10 / 2x + 1^ 2 remember

the key to successfully applying the

quotient rule is to First identify you V

second find their derivatives third

substitute into the quotient rule

formula and finally simplify the

resulting expression

How to Differentiate Rational Functions ((𝟑𝒙^𝟐+𝟓)/(𝟐𝒙+𝟏)) Step-by-Step Tutorial

In this video, we dive into the differentiation of rational functions using the quotient rule. Follow along as we break down each step to find the derivative of ((𝟑𝒙^𝟐+𝟓)/(𝟐𝒙+𝟏)). Master calculus with our clear and concise explanations, and boost your math skills today!
#Calculus #Differentiation #MathTutorial #QuotientRule #LearnMath


let's find the derivative of 2x^2 + 3x +

1 using the definition of the derivative

the definition of the derivative states

that the derivative is equal to the

Limit as H approaches Z of f ofx + hus F

ofx all divided H let's start by finding

F ofx + H we replace every X in our

original function with x + H so

we get 2 * x + h^ 2 + 3 * x + H + 1 now

let's expand this expression first let's

expand the squared term x + h^2 becomes

x^2 + 2x H + h^ 2 multiplying by 2 gives

us 2 x^2 + 4 x h + 2 h^ 2 then we add 3x

+ 3 H + 1 now we can substitute this and

our original function into the

derivative definition we subtract F ofx

which is 2x^2 + 3x + 1 and divide

everything by H let's simplify what's in

the numerator the 2x^2 terms cancel out

the ones cancel out and we're left with

4xh + 2 h^2 + 3 H in the numerator we

can factor out H from the numerator Ator

this gives us H * the quanti 4x + 2 H +

3 H the H is cancel out now we just have

the limit as H approaches Z of 4x + 2 H

+ 3 finally as H approaches 0 2 H

approaches 0 leaving us with 4x + 3 this

is our derivative let's visualize what

this means geometrically here's our

original function 2x^2 + 3x + 1 and at

any point the derivative for X +3 gives

us the slope of the tangent line at that

point in conclusion using the definition

of the derivative we found that the

derivative of 2x^2 + 3x + 1 is 4x + 3

this tells us the instantaneous rate of

change of our function at any point x

Understanding the Derivative of 2x^2+3x+1 Using First Principles

Master the fundamentals of calculus by deriving the exponential function \(2x^2 + 3x + 1\) from first principles. Explore our step-by-step explanations and examples to deepen your understanding of derivatives and exponential functions.

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


let's solve this challenging derivative

problem involving inverse cosine we'll

break it down step by step to make it

more manageable first let's simplify our

notation by letting y equal the inverse

cosine of 4X Cub - 3x then we'll make a

clever substitution let x equal cosine

Theta when we substitute this into our

equation we get y equal the inverse

cosine of 4 cosine Cub thet - 3 cosine

Theta this is where we can use a very

important trigonometric identity for

cine Cub thet - 3 cosine thet = cosine

of 3 Theta therefore our equation

simplifies to y equals the inverse

cosine of cine 3 Theta the inverse

cosine and cosine functions cancel out

leaving us with y = 3 Theta now remember

that Theta is the inverse cosine of x so

we can write y = 3 * the inverse cosine

of x to find the derivative we can use

the chain rule first let's take the

derivative of 3 * the inverse cosine of

x the derivative of inverse cosine is -1

over theun of 1 - x^2 therefore when we

multiply by 3 we get -3 over the sare <

TK of 1 - x^2 and this gives us our

final answer the derivative of the

inverse cosine of 4X Cub - 3x = -3 over

the < TK of 1 - x^2 this problem

demonstrates the power of trigonometric

substitution in calculus by making a

clever substitution we were able to

transform a complex derivative into a

much simpler one

Mastering Inverse Trigonometric Derivatives [𝑐𝑜𝑠^(−1)(4𝑥^3−3𝑥)] Step-by-Step Guide

Question:
Evaluate the derivative [𝑐𝑜𝑠^(−1)(4𝑥^3−3𝑥)].

Join us as we dive into the process of evaluating derivatives involving inverse trigonometric functions. This video will guide you through each step of finding the derivative of [𝑐𝑜𝑠^(−1)(4𝑥^3−3𝑥)], breaking it down with clear explanations. Perfect for students tackling calculus!
#Calculus #InverseTrig #Derivatives #Mathematics #LearningMath


welcome today we'll find the derivative

of the inverse tangent of 1 + x over 1 -

x let's break this down step by step

let's start by letting y equal our

expression this will help us work

through the problem systematically now

there's a very useful identity for

inverse tangent that we can apply here

the identity states that the inverse

tangent of you plus v over 1 minus UV

equals the sum of inverse tangent of U

plus inverse tangent of v in our case if

we let you equal 1 and V equal x we can

see that our expression matches this

form perfectly therefore we can write

our expression as inverse tangent of 1 +

inverse tangent of X now we know that

inverse tangent of 1 = < / 4 this is

because tangent of < / 4 = 1 now that we

have simplified our expression we can

take the derivative the derivative of a

constant pi over 4 is zero and the

derivative of inverse tangent of X is 1/

1 + x^2 therefore the derivative of

inverse tangent of 1 + x/ 1 - x = 1 / 1

+ x^2 let's visualize this result the

blue curve shows our original function

and the green curve shows its derivative

Understanding the Derivative of Inverse Tangent Functions [𝑡𝑎𝑛^(−1)((1+𝑥)/(1−𝑥))]

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

Dive into the world of calculus as we explore the step-by-step process of finding the derivative of [𝑡𝑎𝑛^(−1)((1+𝑥)/(1−𝑥))]. Our detailed explanation will help you understand trigonometric identities and calculus techniques, enhancing your math skills. Perfect for students and math enthusiasts!
#Calculus #MathTutorial #InverseTangent #Differentiation #Trigonometry


let's solve this calculus problem step

by step we need to find the derivative

of Ln sin of 2x first let's identify the

structure of our function we have an

outer natural log function and inside

that we have S of 2x this means we'll

need to use the chain rule let's recall

the chain rule the derivative of a

composite function is the derivative of

the outer function evaluated at the

inner function times the derivative of

the inner function now let's find the

derivative of the outer function the

derivative of natural log is 1 / U so in

our case that's 1 / sin of 2x next let's

find the derivative of the inner

function the derivative of s is cosine

and when we have 2X inside we need to

multiply by the derivative of 2x which

is 2 so we get cosine of 2 x * 2 now

let's put it all together we multiply 1

/ s of 2x * cine of 2x * 2 we can

rearrange this the two can be moved to

the front and cosine over s is the

definition of cotangent and there's our

final answer the derivative of Ln sin of

2x = 2 cent of 2x let's review the key

steps we used first we identified the

composite function structure then we

applied the chain rule next we found the

derivatives of both the outer and inner

functions finally we simplified our

expression using the definition of coent

Understanding the Derivative of [(ln(sin2x))]   Step by Step Explanation

Question: Find the derivative of [(ln(sin2x))]
In this video, we break down the process of finding the derivative of (ln(sin2x))  using the chain rule and trigonometric identities. Follow along for a clear and concise explanation to strengthen your calculus skills!
#hsc #hschighermath #1stpaper #hschighermath1stpaper #englishversion #cpmrana #Calculus #Derivative #Trigonometry #ChainRule #MathTutorial


let's solve this derivative step by step

we're going to find the derivative of

the cube root of 3x^2 + 1 the first step

is to rewrite our cube root in

exponential form the cube root is the

same as raising something to the 1/3

power so we can rewrite this as 3x^2 + 1

raised to the 13 power for step two

we'll use the chain rule since we have a

composite function the chain rule tells

us that we multiply the derivative of

the outer function by the derivative of

the inner function for step three let's

focus on the power rule part when we

have something raised to the 1/3 power

and we differentiate it we multiply by

1/3 and reduce the power by one that's

why we see the 1/3 minus one in the

exponent for step four we need to find

the derivative of the inner function

3x^2 + 1 the derivative of x^2 is 2x so

the derivative of 3x^2 is 6X and the

derivative of 1 is 0er for step five we

multiply everything together we have 1/3

* the expression with the -23 power *

6X finally we can simplify this 1/3 * 6X

is 2x so our final answer is 2x * 3x^2 +

1 raed to the -2/3 power let's verify

this makes sense our original function

was a cube root which has a positive

exponent of 1/3 after taking the

derivative we expect the power to

decrease by 1 giving us -2/3 which is

exactly what we got the X term in front

comes from differentiating the inside of

our cube root this confirms our answer

is reasonable

9.3-1.1 How to Differentiate Cubic Root Functions (∛(3𝑥^2+1)) Step-by-Step Math Explanation.

Description:
In this video, we break down the differentiation of cubic root functions using the chain rule and power rule. Learn how to find the derivative of ∛(3𝑥^2+1)  with clear, step-by-step explanations. Perfect for calculus students looking to master derivatives!
#Calculus #Differentiation #MathTutorial #ChainRule #PowerRule #cpmrana


let's solve this differentiation problem

step by step we need to find the

derivative of x * the square < TK of x^2

+ a 2 first let's identify that this is

a product of two functions we have = X

and V = the < TK of x^2 + a 2 this means

we'll need to use the product rule

applying the product rule we get x * the

derivative of the square root term plus

the square root term times the

derivative of x now for the first term

we need to use the chain rule to

differentiate the square root the

derivative of square root of U is 1 / 2

sare < TK of U * the derivative of U

applying this to our problem we get x/ 2

< TK of x^2 + a squar times the

derivative of what's inside the square

root plus the root term the derivative

of x^2 + a s is simply 2x since the

derivative of a squar is 0o so we get x

* 2x over 2 < TK of x^2 + a 2 plus the

square root term now let's simplify the

numerator becomes x^2 and we can write

the second term as x^2 + a 2 over the

square < TK of x^2 + a 2 finally combine

in these terms under a common

denominator we get 2x^2 + a 2 over

theare < TK of x^2 plus a squar and this

is our final answer we use both the

product rule and chain rule to

differentiate this expression and

arrived at this simplified form

How to Differentiate (𝑥√(𝑥^2+𝑎^2 )) : Step-by-Step Guide

Learn how to differentiate the expression (𝑥√(𝑥^2+𝑎^2 )) using the product and chain rules. This video breaks down each step with clear explanations to help you master the differentiation process. Perfect for students and anyone looking to improve their calculus skills.
#Calculus
#MathTutorial
#Differentiation
#ProductRule
#ChainRule


let's learn about the product rule for

derivatives this rule tells us how to

find the derivative of a product of two

functions let's start with two functions

u and v where both are functions of X we

want to find the derivative of their

product let's visualize this

geometrically imagine you and V

represent the sides of a rectangle the

product U * V represents the area of

this rectangle when X changes both u and

v change causing the area to change in

two ways first you changes while V stays

constant and then V changes while you

stays constant the total change in area

is the sum of these two changes this

gives us the product rule the derivative

of U * V equal U * the derivative of V

plus v * the derivative of U let's try

an example let's find the derivative of

x^2 * X Cub here U is x^2 and V is X

cubed let's apply the product rule First

We Take U which is x^2 and multiply it

by the derivative of V which is

3x^2 then we add V which is X Cub * the

derivative of U which is 2x simplifying

this gives us 5x to 4th power remember

the the product rule states that when

differentiating a product of two

functions we multiply each function by

the derivative of the other function and

add the results find

derivative using product rule let's

solve this calculus problem step by step

we need to find the derivative of x^2 *

natural log of x to solve this we'll

need to use the product rule the product

rule states that when differentiating

the product of two functions we multiply

the first function by the derivative of

the second plus the second function

times the derivative of the first in our

case let's identify that U of X is X2

and V of X is natural log of x so

applying the product rule we get x^2 *

the derivative of natural log of x plus

natural log of x * the derivative of x^2

now let's evaluate each derivative

separately first the derivative of

natural log of x is

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

substituting these derivatives back into

our equation we get x^2 *

1X Plus natural log of x *

2x now let's simplify each term the

first term X2 * * 1 /x simplifies to

just X the second term natural log of x

* 2x becomes 2x natural log of x for our

final step we can factor out X from both

terms giving us x * thean 1 + 2 natural

log of x and this is our final answer to

verify this is correct we can see that

we've properly applied the product rule

simplified our terms and fact facted our

result the derivative of x^2 * natural

log of x is x * thean 1 + 2 natural log

of x

Product Rule for Differentiation Explained with  Example.

Product Rule for Differentiation Explained with Easy Examples | Step-by-Step Guide
Learn the Product Rule for Differentiation in this step-by-step tutorial! 🎓 We explain the formula, break it down with examples, and show you how to solve problems involving the product of two functions.
📌 Topics Covered in This Video:
What is the Product Rule?
Formula and explanation
Step-by-step example solving
Example Problem Included:
We calculate the derivative of x^2 lnx using the product rule.
👉 Perfect for students learning calculus or preparing for exams!

💡 Don’t forget to LIKE, SHARE, and SUBSCRIBE for more math tutorials!
#Calculus #ProductRule #Differentiation #MathTutorial #LearnMath #BengaliEducation #MathMadeEasy #MathTips


let's solve this calculus problem step

by step using the quotient rule we need

to find the derivative of x to the N

divided natural log of x first let's

recall the quotient rule formula 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 this case U

equal x to the n and V equal natural log

of x now let's find the derivatives of u

and v the derivative of x to the N is NX

the

nus1 and the derivative of natural log

of x is 1 /x Let's substitute these into

our quotient rule formula we get natural

log x * NX

n-1 - x n * 1X all over natural log x^2

now let's simplify the numerator in the

second term X the N / X simplifies to X

N -1 for our final step we can factor

out x to the nus 1 from both terms in

the numerator this gives us x to the nus

1 * the quantity n natural log x -1 all

over natural log x^2 and this is our

final answer answer we've successfully

Mastering Derivatives: Differentiating [(x^n /lnx)]  Using the Quotient Rule

Join us as we delve into the world of calculus and learn how to differentiate the expression [(x^n /lnx)]   using the quotient rule. Perfect for students preparing for calculus exams or anyone looking to strengthen their understanding of mathematics.
#Calculus #MathTutorial #Differentiation #QuotientRule #AdvancedMath


let's explore the fundamental equations

of motion in kinematics these equations

help us understand how objects move

under constant acceleration the first

equation relates final velocity initial

velocity acceleration and time V equal U

+ a t where V is the final velocity U is

the initial velocity uh is acceleration

and T is time the second equation gives

us a relationship between velocities

without needing time V ^2 = u^2 + 2 a s

this is particularly useful when we know

the displacement s but don't know the

time taken the third equation helps us

find displacement s = u t + 12 a t sared

this shows us that displacement has two

parts the part due to initial velocity

UT and the part due to acceleration 12 a

t^2 the fourth equation uses average

velocity s equal the average of U + V *

T this tells us that for constant

acceleration we can use the average of

initial and final velocities to find

displacement and finally the fifth

equation is for constant velocity motion

when velocity doesn't change S equals VT

this is the simplest case where

acceleration is zero together these

equations form the foundation of

kinematics allowing us to solve any

problem involving motion under constant

acceleration remember that each equation

is useful in different situations

depending on what information you have

and what you're trying to find

Equation of motion

“Understanding Kinematic Equations: A Step-by-Step Guide”
Description:
Dive into the world of kinematics with our detailed explanation of key motion equations! In this video, we break down the fundamental kinematic equations that describe motion under constant acceleration, including:
These equations are essential for understanding how objects move. Whether you’re a student or just curious about physics, this video is designed to help you grasp the concepts clearly.
Join us for a clear, step-by-step explanation and enhance your understanding of motion!
#Kinematics #Physics #MotionEquations #Acceleration #Velocity #ScienceEducation #PhysicsTutorial #Displacement #LearnPhysics


let's solve a challenging derivative

problem involving inverse trigonometric

functions our task is to find the

derivative of sin inverse of 2x * the <

TK of 1 - x^2 the key to solving this

problem is to use substitution to

simplify the expression let's start by

letting y equal sin inverse of 2x * < TK

1 - x^2 now comes the clever part we'll

make a

substitution let X = sin Theta this

means theta equals sin inverse of X

let's substitute this into our

expression when we replace x with sin

Theta we get sin inverse of 2 sin thet *

the < TK of 1 - sin^2 th Now 1 - sin^2

is equal to cine 2 this is a fund Al

trigonometric identity the square root

of cosine 2 thet is simply cosine Theta

so our expression simplifies to sin

inverse of 2 sin Theta cine Theta here's

where another key identity comes in 2

sin Theta cine Theta = sin of 2 thet

this is the double angle formula for S

when we have sin inverse of s of

something they cancel out leaving us

with Just 2 Theta

remember thata is sin inverse of X so

our final expression for y is 2 * sin

inverse of X now we can find the

derivative the derivative of y with

respect to x equals the derivative of 2

sin inverse of X using the chain rule

this equals 2 * the derivative of sin

inverse of X which equal 2 / the sare <

TK of 1 - x^2 therefore the derivative

of sin inverse of 2x * the < TK of 1 -

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

How to Find the Derivative of Inverse Sine Function [𝑠𝑖𝑛^(−1){2𝑥 √(1−𝑥^2 )}] Step-by-Step Guide

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

In this video, we dive into the calculus problem of finding the derivative of an inverse sine function, specifically [𝑠𝑖𝑛^(−1){2𝑥 √(1−𝑥^2 )}] .. Follow along as we show you a two-column approach, explaining each mathematical step alongside a detailed explanation. This clear and concise walkthrough will help you understand the differentiation process and apply it to similar problems. Perfect for students looking to strengthen their calculus skills!
#Calculus #Derivative #InverseSine #MathTutorial #StepByStep


let's solve this derivative step by step

using the chain rule we'll see how to

differentiate cosine to the 4th power of

theta to solve this we'll use the chain

rule the chain rule tells us that when

we have a composite function which means

a function inside another function we

multiply the derivative of the outer

function by the derivative of the inner

function in our case we have cosine of

theta as our inner function and the

fourth power as our outer function let's

identify these parts first let's find

the derivative of the outer function the

derivative of U to the 4th power with

respect to U is 4 U cubed substituting

back our inner function we get 4 *

cosine cubed of theta next we need the

derivative of the inner function the

derivative of cosine Theta is sin Theta

now according to the chain rule we

multiply these parts together so we have

for cine cubed thet * sin Theta this

gives us our final answer -4 cosine

cubed Theta sin Theta let's visualize

why this makes sense here's the graph of

cosine to the 4th power and here's its

derivative that we just found notice how

the derivative is zero at the Peaks and

troughs of the original function and

steepest at the middle points just as we

would expect and that's how we find the

derivative of cosine to the 4th power of

theta using the chain rule remember when

dealing with composite functions always

identify your inner and outer functions

and apply the chain rule step by step

9.3-1.5 Understanding Derivatives Differentiating (𝑐𝑜𝑠^4𝜃)  with the Chain Rule

Learn how to differentiate (𝑐𝑜𝑠^4𝜃)  step-by-step using the chain rule and power rule. We’ll break down the process to help you understand the math behind it. Perfect for students and anyone looking to improve their calculus skills!
#Calculus #MathTutorial #Differentiation #ChainRule #MathHelp #Trigonometry #cpmrana


let's solve this complex trigonometric

differentiation problem step by step

we'll start with finding the derivative

of this inverse tangent expression let's

begin by letting y equal our expression

we'll start with Y = inverse tangent of

a cosine x - B sinx over b cosin x + a

sinx our first step is to simplify this

fraction we can factor out B from both

numerator and denominator giving us this

expression the B's cancel out in the

fraction simplifying our expression to

this form now comes the clever part

we'll multiply both numerator and

denominator by 1 / cosine X this will

help us introduce tangent X terms

remember that sinx over cine X is

tangent X this gives us a simpler form

this is a key step we can recognize this

as the tangent of a difference formula

when we have tangent inverse of uus V

over 1 + UV it equals tangent inverse

uus tangent inverse V applying this

formula where U = / B and V equal

tangent X we get tangent inverse of A

over B minus tangent inverse of tangent

X since t tent inverse of tangent x = x

for our domain we can simplify this to

tangent inverse of a over B- X now we

can find

dydx the derivative of tangent inverse

of a constant is0 and the derivative ofx

is -1 therefore the derivative of our

original expression equals -1

Arctan Trigonometric Derivatives [𝑡𝑎𝑛^(−1)((𝑎 cos𝑥−𝑏 sin𝑥)/(𝑏 cos𝑥+𝑎 sin𝑥 ))] Step-by-Step

Question:
Prove that the derivative of the function [𝑡𝑎𝑛^(−1)((𝑎 cos𝑥−𝑏 sin𝑥)/(𝑏 cos𝑥+𝑎 sin𝑥 ))]  is  -1.

Dive into solving arctan trigonometric derivatives step-by-step! This video covers the intricate process of differentiating the expression [𝑡𝑎𝑛^(−1)((𝑎 cos𝑥−𝑏 sin𝑥)/(𝑏 cos𝑥+𝑎 sin𝑥 ))]. Follow along to master how each step unfolds with clear explanations.

#Calculus #Trigonometry #Arctan #MathTutorial #Differentiation #MathHelp


let's solve this complex derivative

problem step by step we need to find the

derivative of x^2 * log base a of x + 7

e to x * cine X first let's break this

into two parts using the sum rule of

derivatives we'll differentiate x^2 log

base uh of X separately from 7 e to x

cosine X let's tackle the first part 4X

squ log Bas of X we'll use the product

rule remember the product rule states

that the derivative of U * v = u Prime *

V + U * V Prime this gives us x^2 * the

derivative of log base a of X Plus log

base a of x * the derivative of

x^2 now for the derivative of log base a

of X we can use the change base formula

log Bas of xal log Bas e of x / log base

e of a the derivative of natural log is

1 /x for the second part 7 e to x cosine

X will again use the product rule this

gives us 7 * e to x * sinx + cine x * e

to the

X now let's combine our results the

final answer is x * log Bas of e + 2 log

Bas of x + 7 e to x * the qu cosine x -

sin

x let's verify this is correct by

checking our work we use the sum rule to

separate the terms the product rule

twice and the chain rule for the

logarithm each step follows from the

fundamental rules of

differentiation and that's how we find

this complex derivative remember to

break down complicated problems into

smaller manageable Parts using the basic

rules of

differentiation thank you for watching

Finding the Derivative of (𝑥^2 log_𝑎^𝑥+7𝑒^𝑥  cos𝑥)  Using the Product Rule | Step-by-Step.

In this video, learn to differentiate  (𝑥^2 log_𝑎^𝑥+7𝑒^𝑥  cos𝑥)  using the Product Rule in calculus.

What You'll Learn:
- Definition of the Product Rule
- Step-by-step differentiation of  (𝑥^2 log_𝑎^𝑥+7𝑒^𝑥  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 #hsc


let's solve this complex differentiation

problem step by step we'll use

substitution and trigonometric

identities to simplify this seemingly

difficult expression our first step is

to make a substitution to simplify this

expression let's substitute X = Tan t

which means T = inverse tangent of X now

let's substitute this into our original

expression we get tan inverse of the

square < TK of 1 + t^ 2 t - 1 all over

Tan t here's where we can use our first

trigonometric identity we know that 1 +

tan^ 2 T = SEC 2 T making this

substitution gives us tan inverse of

square < TK of secant 2

tus1 over Tan t next we can simplify the

square root of secant squ T to just

secant T this gives us tan inverse of

secant tus1 / Tan t now let's use

another identity remember that secant T

= 1 / cosine T and tangent T = sin t/

cine T substituting these and gives us

this fraction through algebraic

manipulation this simplifies to tan

inverse of 1 - cosine t / sin T here's

where we use another key identity 1 -

cosine T = 2 sin s t / 2 and sin T = 2

sin T / 2 * cine T / 2 substituting

these and gives us this expression the

sin t/ 2 terms cancel out leaving us

with just t/2 remember that t equals

inverse tangent of X so we can write

this as 12 * inverse tangent of X now we

can use the chain rule the derivative of

inverse tangent of X is 1 1 + x^2 and

multiplying by 1/ 12 gives us our Final

Answer 1 over 2 * 1 + x^2 and that's our

solution through careful substitution

and use of trigonometric identities we

transformed a complex expression into a

simple one

Understanding the Derivative of Arctan Functions [𝑡𝑎𝑛^(−1)   (√(1+𝑥^2 )−1)/𝑥] Step-by-Step

Question
Find the derivative of the function [𝑡𝑎𝑛^(−1)   (√(1+𝑥^2 )−1)/𝑥].

Dive into this detailed tutorial where we explore the derivative of the function [𝑡𝑎𝑛^(−1)   (√(1+𝑥^2 )−1)/𝑥]. Follow along as we break down each step to simplify and differentiate this expression using trigonometric substitutions and identities. Perfect for students looking to enhance their calculus skills!
#Calculus #MathTutorial #Differentiation #ArctanDerivatives #TrigonometricSubstitution


let's learn how to find the derivative

of log base 10 of X we'll go through

this step by step to understand the

process clearly the first key step is to

convert this logarithm using the change

of Base formula this formula states that

log base uh of x equals the natural log

of x / the natural log of a applying

this to our problem log base 10 of X =

natural log of x time log base 10 of e

notice that log base 10 of e is a

constant now we can rewrite our

derivative problem we're differentiating

the product of natural log of x and the

constant log base 10 of e since log base

10 of e is constant we can Factor it out

of the derivative this gives us log base

10 of e * the derivative of natural log

of x now we use a key fact the

derivative of natural log of x is 1 /x

applying this we get log base 10 of e *

1X our final answer can be written as

log base 10 of

e/x this is the derivative formula for

log base 10 of X let's visualize what

this derivative means on a graph the

function log base 10 of X is shown in

blue and its derivative log base 10 of e

overx is shown in red notice how the

derivative is always positive but

decreases as X increases showing that

log base 10 of X grows more slowly as X

gets larger

Derivative of Logarithms [𝒍𝒐𝒈_𝟏𝟎(𝒙)] Step by Step Explanation

Question: Find the derivative of [𝒍𝒐𝒈_𝟏𝟎(𝒙) ]
In this video, we break down the process of differentiating [𝒍𝒐𝒈_𝟏𝟎(𝒙)]  into simple, easy-to-follow steps. Learn how to apply the change of base formula, differentiate products, and simplify the result. Perfect for students and anyone looking to master calculus concepts!
#Calculus #Logarithms #Differentiation #MathTutorial #Mathematics #MathHelp