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

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

& 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{aligned}$$ Therefore on differentiating $$f\left( x \right)$$ with respect to $$x$$, we get $$\begin{aligned} & \dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( {{\sin }^{-1}}x \right) \\\ & =\dfrac{1}{\sqrt{1-{{x}^{2}}}}.........(2) \end{aligned}$$ Now in order to calculate $$\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{aligned} & 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{aligned}$$ Therefore on differentiating $$g\left( x \right)$$ with respect to $$x$$, we get $$\begin{aligned} & \dfrac{d}{dx}g\left( x \right)=\dfrac{d}{dx}\left( 2{{\sin }^{-1}}x \right) \\\ & =\dfrac{2}{\sqrt{1-{{x}^{2}}}}.......(3) \end{aligned}$$ Now on substituting the values in equation (2) and equation (3) in (1), we get $$\begin{aligned} & \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{aligned}$$ Therefore we get that the derivative of the function $${{\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$$.