Question

Find the derivative of \[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)\]with respect to \[{{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)\]is
A.\[\dfrac{1}{\sqrt{1-{{x}^{2}}}}\]
B.\[\dfrac{3}{\sqrt{1-{{x}^{2}}}}\]
C.\[3\]
D.\[\dfrac{1}{3}\]

Answer 1

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

Complete step-by-step answer:
Let us now take \[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)\]. . . . . . . . . . . . . . . . . . . . . . . .. . .(1)
Replace the value of x by \[\sin \theta \], then we will get
\[x=\sin \theta \] in equation (1)
\[={{\tan }^{-1}}\left( \dfrac{\sin \theta }{\sqrt{1-{{\sin }^{2}}\theta }} \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(2)
\[={{\tan }^{-1}}\left( \dfrac{\sin \theta }{\sqrt{{{\cos }^{2}}\theta }} \right)\]. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . .(3)
\[={{\tan }^{-1}}\left( \tan \theta \right)\]
\[=\theta \]
We have assumed that \[x=\sin \theta \]
So \[\theta ={{\sin }^{-1}}x\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(4)
Let us now take \[{{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)\]. . . . . . . . . . . . . . . . . . . . . . . .(5)
Replace the value of x by \[\sin \theta \], then we will get
\[x=\sin \theta \] in equation (5)
\[={{\sin }^{-1}}\left( 3\sin \theta -4{{\sin }^{3}}\theta \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . .(6)
\[={{\sin }^{-1}}\left( \sin 3\theta \right)\]. . . . . . . . . . . . . . . . . . .. .. .. .. .. . .(7)
\[=3\theta \]
We have assumed that \[x=\sin \theta \]
So \[\theta ={{\sin }^{-1}}x\]
\[=3{{\sin }^{-1}}x\]. . . . . . . . . . . . . . .. . . . . . .(8)
So we have to find the derivative of \[{{\sin }^{-1}}x\]with respect to \[3{{\sin }^{-1}}x\]
\[\dfrac{d}{dx}\left( {{\sin }^{-1}}x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}\]
\[\dfrac{d}{dx}\left( 3{{\sin }^{-1}}x \right)=\dfrac{3}{\sqrt{1-{{x}^{2}}}}\]
So, the ratio of derivative \[{{\sin }^{-1}}x\] with respect to \[3{{\sin }^{-1}}x\]is \[\dfrac{1}{3}\]
So the correct option is option (D)
Note:The derivative of \[{{\sin }^{-1}}x\] is given by \[\dfrac{d}{dx}\left( {{\sin }^{-1}}x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}\]. 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 \[{{\sin }^{-1}}x\] and \[{{\tan }^{-1}}x\],but if we does not use trigonometric functions to do the sum the answer process is very big. So better to use trigonometric functions