Question

The derivative of ${{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}-1}{x} \right)$ w. r. t ${{\tan }^{-1}}\left( \dfrac{2x\sqrt{1-{{x}^{2}}}}{1-2{{x}^{2}}} \right)$ at $x=0$ is
A. $\dfrac{1}{4}$
B. $\dfrac{1}{8}$
C. $\dfrac{1}{2}$
D. 1

Answer 1

Hint: For the above question we will first simplify the given inverse trigonometric function by substituting the value of $'x'$ to the other trigonometric ratios and then we will differentiate it separately and take its ratio. We will get the required derivative. We will substitute $x=\tan \theta$ in ${{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}-1}{x} \right)$ and $x=\sin \theta$ in ${{\tan }^{-1}}\left( \dfrac{2x\sqrt{1-{{x}^{2}}}}{1-2{{x}^{2}}} \right)$.

Complete step-by-step answer:

We have been asked to find the derivative of ${{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}-1}{x} \right)$ with respect to ${{\tan }^{-1}}\left( \dfrac{2x\sqrt{1-{{x}^{2}}}}{1-2{{x}^{2}}} \right)$ at $x=0$.
Let $u={{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}-1}{x} \right)$.
Now, substitute $x=\tan \theta$.
$\Rightarrow u={{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{\tan }^{2}}\theta }-1}{\tan \theta } \right)$
Since, we know the identity $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta$.
$\begin{aligned} & \Rightarrow u={{\tan }^{-1}}\left( \dfrac{\sqrt{{{\sec }^{2}}\theta }-1}{\tan \theta } \right) \\\ & ={{\tan }^{-1}}\left( \dfrac{\sec \theta -1}{\tan \theta } \right) \\\ & ={{\tan }^{-1}}\left( \dfrac{\dfrac{1}{\cos \theta }-1}{\dfrac{\sin \theta }{\cos \theta }} \right) \\\ & ={{\tan }^{-1}}\left( \dfrac{\dfrac{1-\cos \theta }{\cos \theta }}{\dfrac{\sin \theta }{\cos \theta }} \right) \\\ & ={{\tan }^{-1}}\left( \dfrac{1-\cos \theta }{\sin \theta } \right) \\\ \end{aligned}$
On using the identity $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$ and $1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}$, we get,
$\begin{aligned} & ={{\tan }^{-1}}\left( \dfrac{2{{\sin }^{2}}\dfrac{\theta }{2}}{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}} \right) \\\ & ={{\tan }^{-1}}\left( \tan \dfrac{\theta }{2} \right) \\\ \end{aligned}$
We know that ${{\tan }^{-1}}\tan x=x$
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}-1}{x} \right)={{\tan }^{-1}}\left( \tan \dfrac{\theta }{2} \right)=\dfrac{\theta }{2}$
Since, $x=\tan \theta$
$\begin{aligned} & \Rightarrow \theta ={{\tan }^{-1}}x \\\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}-1}{x} \right)=\dfrac{{{\tan }^{-1}}x}{2} \\\ \end{aligned}$
So, we have $u=\dfrac{{{\tan }^{-1}}x}{2}$
Differentiating both side with respect to x and using formula $\dfrac{d}{dx}{{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}}$, we get,
$\dfrac{du}{dx}=\dfrac{1}{2\left( 1+{{x}^{2}} \right)}..........\left( 1 \right)$
Again, let $v={{\tan }^{-1}}\left( \dfrac{2x\sqrt{1-{{x}^{2}}}}{1-2{{x}^{2}}} \right)$
On substituting $x=\sin \theta$, we get,
$\begin{aligned} & v={{\tan }^{-1}}\left( \dfrac{2\sin \theta \sqrt{1-{{\sin }^{2}}\theta }}{1-2{{\sin }^{2}}\theta } \right) \\\ & ={{\tan }^{-1}}\left( \dfrac{2\sin \theta \sqrt{{{\cos }^{2}}\theta }}{1-2{{\sin }^{2}}\theta } \right) \\\ & ={{\tan }^{-1}}\left( \dfrac{2\sin \theta \cos \theta }{1-2{{\sin }^{2}}\theta } \right) \\\ \end{aligned}$
Since, we know that $2\sin A\cos A=\sin 2A$ and $1-2{{\sin }^{2}}A=\cos 2A$.
$\begin{aligned} & ={{\tan }^{-1}}\left( \dfrac{\sin 2\theta }{\cos 2\theta } \right) \\\ & ={{\tan }^{-1}}\left( \tan 2\theta \right) \\\ \end{aligned}$
Since, we know that ${{\tan }^{-1}}\tan x=x$
$\Rightarrow {{\tan }^{-1}}\left( \tan 2\theta \right)=2\theta$
Since, we have supposed $x=\sin \theta$. So, on taking sine inverse both sides we get ${{\sin }^{-1}}x=\theta$.
Substituting the value of $\theta$ we get,
$v=2{{\sin }^{-1}}x$
Differentiating the above equation with respect to x, we get,
$\dfrac{dv}{dx}=\dfrac{2}{\sqrt{1-{{x}^{2}}}}............\left( 2 \right)$
Since, the derivative of $\sin x=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$.
On dividing (1) by (2), we get,
$\begin{aligned} & \dfrac{du}{dx}.\dfrac{dx}{dv}=\dfrac{1}{2\left( 1+{{x}^{2}} \right)}.\dfrac{\sqrt{1-{{x}^{2}}}}{2} \\\ & \Rightarrow \dfrac{du}{dv}=\dfrac{1}{4}\dfrac{\sqrt{1-{{x}^{2}}}}{\left( 1+{{x}^{2}} \right)} \\\ \end{aligned}$
Now, at $x=0$
$\dfrac{du}{dv}=\dfrac{1}{4}\dfrac{\sqrt{1-0}}{1+0}=\dfrac{1}{4}$
Therefore, the derivative of ${{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}-1}{x} \right)$ with respect to ${{\tan }^{-1}}\left( \dfrac{2x\sqrt{1-{{x}^{2}}}}{1-2{{x}^{2}}} \right)$ at $x=0$ is equal to $\dfrac{1}{4}$. So, the correct option of the given question is option A.

Note: Remember the point that if we have the expression $\sqrt{1+{{x}^{2}}}$, then we must have to put $x=\tan \theta$ and if we have the expression like $\sqrt{1-{{x}^{2}}}$, then we must substitute $x=\sin \theta \ or\ \cos \theta$ in order to simplify the given function.
Also, remember that if we have given two functions and we have been asked to find derivatives of one function with respect to another then we will separately find its derivative and then take its ratio.