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Question: Find the derivative of \[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)\]with respect to...

Find the derivative of tan⁡−1(x1−x2){{\tan }^{-1}}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)with respect to sin⁡−1(3x−4x3){{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)is
A.11−x2\dfrac{1}{\sqrt{1-{{x}^{2}}}}
B.31−x2\dfrac{3}{\sqrt{1-{{x}^{2}}}}
C.33
D.13\dfrac{1}{3}

Explanation

Solution

Hint: Firstly we have to get thought that we have to replace by sin⁡θ\sin \theta and then we have to proceed. We must know the basic trigonometric functions that is tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta }and sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta and the concept of inverse function that is if x=sin⁡θx=\sin \theta then θ=sin⁡−1x\theta ={{\sin }^{-1}}x

Complete step-by-step answer:
Let us now take tan⁡−1(x1−x2){{\tan }^{-1}}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right). . . . . . . . . . . . . . . . . . . . . . . .. . .(1)
Replace the value of x by sin⁡θ\sin \theta , then we will get
x=sin⁡θx=\sin \theta in equation (1)
=tan⁡−1(sin⁡θ1−sin⁡2θ)={{\tan }^{-1}}\left( \dfrac{\sin \theta }{\sqrt{1-{{\sin }^{2}}\theta }} \right). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(2)
=tan⁡−1(sin⁡θcos⁡2θ)={{\tan }^{-1}}\left( \dfrac{\sin \theta }{\sqrt{{{\cos }^{2}}\theta }} \right). . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . .(3)
=tan⁡−1(tan⁡θ)={{\tan }^{-1}}\left( \tan \theta \right)
=θ=\theta
We have assumed that x=sin⁡θx=\sin \theta
So θ=sin⁡−1x\theta ={{\sin }^{-1}}x. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(4)
Let us now take sin⁡−1(3x−4x3){{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right). . . . . . . . . . . . . . . . . . . . . . . .(5)
Replace the value of x by sin⁡θ\sin \theta , then we will get
x=sin⁡θx=\sin \theta in equation (5)
=sin⁡−1(3sin⁡θ−4sin⁡3θ)={{\sin }^{-1}}\left( 3\sin \theta -4{{\sin }^{3}}\theta \right). . . . . . . . . . . . . . . . . . . . . . . . . .(6)
=sin⁡−1(sin⁡3θ)={{\sin }^{-1}}\left( \sin 3\theta \right). . . . . . . . . . . . . . . . . . .. .. .. .. .. . .(7)
=3θ=3\theta
We have assumed that x=sin⁡θx=\sin \theta
So θ=sin⁡−1x\theta ={{\sin }^{-1}}x
=3sin⁡−1x=3{{\sin }^{-1}}x. . . . . . . . . . . . . . .. . . . . . .(8)
So we have to find the derivative of sin⁡−1x{{\sin }^{-1}}xwith respect to 3sin⁡−1x3{{\sin }^{-1}}x
ddx(sin⁡−1x)=11−x2\dfrac{d}{dx}\left( {{\sin }^{-1}}x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}
ddx(3sin⁡−1x)=31−x2\dfrac{d}{dx}\left( 3{{\sin }^{-1}}x \right)=\dfrac{3}{\sqrt{1-{{x}^{2}}}}
So, the ratio of derivative sin⁡−1x{{\sin }^{-1}}x with respect to 3sin⁡−1x3{{\sin }^{-1}}xis 13\dfrac{1}{3}
So the correct option is option (D)
Note:The derivative of sin⁡−1x{{\sin }^{-1}}x is given by ddx(sin⁡−1x)=11−x2\dfrac{d}{dx}\left( {{\sin }^{-1}}x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}. This problem is solved using trigonometric functions. If we did not get the thought that we have to use trigonometric function the we can proceed using derivative formulas of sin⁡−1x{{\sin }^{-1}}x and tan⁡−1x{{\tan }^{-1}}x,but if we does not use trigonometric functions to do the sum the answer process is very big. So better to use trigonometric functions