Question

Differentiate the function \[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)\] with respect to \[{{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\].

Answer 1

Hint: In this question, in order to find differentiation the function \[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)\] with respect to \[{{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\] we will suppose that the function \[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)\] is equal to \[f\left( x \right)\] and the function \[{{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\] is equals to the function \[g\left( x \right)\]. Now the problem is to find the differential of the function \[f\left( x \right)\] with respect to \[g\left( x \right)\]. That is we have to find the \[\dfrac{df\left( x \right)}{dg\left( x \right)}\]. Now since both \[f\left( x \right)\] and \[g\left( x \right)\] are function of \[x\], thus we have \[\dfrac{df\left( x \right)}{dg\left( x \right)}=\dfrac{d}{dx}f\left( x \right)\div \dfrac{d}{dx}g\left( x \right)\].Now in order to find the derivatives \[\dfrac{d}{dx}f\left( x \right)\] and \[\dfrac{d}{dx}g\left( x \right)\] we will substitute \[x=\sin t\] and then solve the problem.

Complete step by step answer:
Let us suppose that the function \[f\left( x \right)\] is given by \[f\left( x \right)={{\tan }^{-1}}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)\].
And the function \[g\left( x \right)\] is given by \[g\left( x \right)={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\].
Now I will have to find the differential of the function \[f\left( x \right)\] with respect to \[g\left( x \right)\].
That is we have to find the value of \[\dfrac{df\left( x \right)}{dg\left( x \right)}\].
Also since both the functions \[f\left( x \right)\] and \[g\left( x \right)\] are function of \[x\], thus we have \[\dfrac{df\left( x \right)}{dg\left( x \right)}=\dfrac{d}{dx}f\left( x \right)\div \dfrac{d}{dx}g\left( x \right)...........(1)\]
So now we have to find the derivatives \[\dfrac{d}{dx}f\left( x \right)\] and \[\dfrac{d}{dx}g\left( x \right)\].
Let us first suppose that \[x=\sin t\].
Now in order to calculate \[\dfrac{d}{dx}f\left( x \right)\], we will substitute the value \[x=\sin t\] in \[f\left( x \right)\]. Then we get
\[\begin{align}
  & f\left( x \right)={{\tan }^{-1}}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right) \\
 & ={{\tan }^{-1}}\left( \dfrac{\sin t}{\sqrt{1-{{\sin }^{2}}t}} \right) \\
 & ={{\tan }^{-1}}\left( \dfrac{\sin t}{\cos t} \right) \\
 & ={{\tan }^{-1}}\left( \tan t \right) \\
 & =t \\
 & ={{\sin }^{-1}}x
\end{align}\]
Therefore on differentiating \[f\left( x \right)\] with respect to \[x\], we get
\[\begin{align}
  & \dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( {{\sin }^{-1}}x \right) \\
 & =\dfrac{1}{\sqrt{1-{{x}^{2}}}}.........(2)
\end{align}\]
Now in order to calculate \[\dfrac{d}{dx}g\left( x \right)\], we will substitute the value \[x=\sin t\] in \[g\left( x \right)\].
Since we know that \[\sin 2t=2\sin t\cos t\], then we get
\[\begin{align}
  & g\left( x \right)={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right) \\
 & ={{\sin }^{-1}}\left( 2\sin t\sqrt{1-{{\sin }^{2}}t} \right) \\
 & ={{\sin }^{-1}}\left( 2\sin t\cos t \right) \\
 & ={{\sin }^{-1}}\left( \sin 2t \right) \\
 & =2t \\
 & =2{{\sin }^{-1}}x
\end{align}\]
Therefore on differentiating \[g\left( x \right)\] with respect to \[x\], we get
\[\begin{align}
  & \dfrac{d}{dx}g\left( x \right)=\dfrac{d}{dx}\left( 2{{\sin }^{-1}}x \right) \\
 & =\dfrac{2}{\sqrt{1-{{x}^{2}}}}.......(3)
\end{align}\]
Now on substituting the values in equation (2) and equation (3) in (1), we get
\[\begin{align}
  & \dfrac{df\left( x \right)}{dg\left( x \right)}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}\div \dfrac{2}{\sqrt{1-{{x}^{2}}}} \\
 & =\dfrac{1}{\sqrt{1-{{x}^{2}}}}\times \dfrac{\sqrt{1-{{x}^{2}}}}{2} \\
 & =\dfrac{1}{2}
\end{align}\]
Therefore we get that the derivative of the function \[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)\] with respect to \[{{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\] is equals to \[\dfrac{1}{2}\].

Note:
In this problem, we cannot directly calculate the derivative of the function \[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)\] with respect to \[{{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\]. Also in this problem we have \[x\in \left[ -\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}} \right]\]. Because when we are substituting the value \[x=\sin t\] in \[g\left( x \right)={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\], we will have \[t\in \left[ -\pi ,\pi \right]\] which implies \[2t\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\]. And therefore we have \[{{\sin }^{-1}}\left( \sin 2t \right)=2t\].