Question

Find the derivative of the following functions w.r to x. i) cos-1(4x3 -3x)

Answer 1

Is the question is cos⁻1(4x³-3x).

Answer 2

Answer:

Find the derivative of the following functions w.r to x

$\rm { \cos}^{ - 1} (4 {x}^{3} - 3x)$

Solution

Let:-

$\rm y = { \cos}^{ - 1} (4 {x}^{3} - 3x)$

let

$\rm \: x = \cos\theta$

so that

$\rm \theta \: = { \cos }^{ - 1} x$

Then

$\rm \: y = { \cos}^{ - 1} (4 { \cos}^{3} \theta - 3 \cos \theta)$

[we know, 4cos³θ - 3cosθ = cos3θ]

$\rm \: y = { \cos }^{ - 1} (\cos 3\theta)$

$\rm[ We \: know \: \: { \cos }^{ - 1} ( \cos \theta) = \theta]$

so,

$\rm \: y = 3 \theta$

$\rm \: y \: = 3 { \cos }^{ - 1} x$

Differentiating both sides, with respect to x

$\rm \dfrac{dy}{dx} = \dfrac{d}{dx} (3 { \cos }^{ - 1} x)$

$\rm \dfrac{dy}{dx} =3. \dfrac{d}{dx} ( { \cos }^{ - 1} x)$

$\rm \dfrac{dy}{dx} = 3. \bigg(- \dfrac{1}{ \sqrt{1 - {x}^{2} } } \bigg)$

$\rm \dfrac{dy}{dx} = - \dfrac{3}{ \sqrt{1 - {x}^{2} } }$

Some Derivative Formulae:-

$\tt \dfrac{d}{dx} ( { \sin }^{ - 1}x) = \dfrac{1}{ \sqrt{1 - {x}^{2} } }$

$\tt \dfrac{d}{dx} ( { \cos}^{ - 1}x) = \dfrac{ - 1}{ \sqrt{1 - {x}^{2} } }$

$\tt \dfrac{d}{dx} ( { \tan}^{ - 1}x) = \dfrac{ 1}{1 + {x}^{2} }$

$\tt \dfrac{d}{dx} ( { \cot}^{ - 1}x) = \dfrac{ - 1}{1 + {x}^{2} }$

$\tt \dfrac{d}{dx} ( { \sec}^{ -1}x) = \dfrac{ 1}{ |x| \sqrt{ {x}^{2} - 1} }$

$\tt \dfrac{d}{dx} ( { \cosec}^{ -1}x) = \dfrac{ - 1}{ |x| \sqrt{ {x}^{2} - 1} }$

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