Question

The derivative of ${{\tan }^{-1}}\left[ \dfrac{\left( \sqrt{1+{{x}^{2}}}-1 \right)}{x} \right]$ with respect to ${{\tan }^{-1}}\left[ \dfrac{\left( 2x\sqrt{1-{{x}^{2}}} \right)}{\left( 1-2{{x}^{2}} \right)} \right]$ at $x=0$ is:
1) $\dfrac{1}{8}$
2) $\dfrac{1}{4}$
3) $\dfrac{1}{2}$
4) $1$

Answer 1

Hint: Here in this question, we have been asked to find the derivative of ${{\tan }^{-1}}\left[ \dfrac{\left( \sqrt{1+{{x}^{2}}}-1 \right)}{x} \right]$ with respect to ${{\tan }^{-1}}\left[ \dfrac{\left( 2x\sqrt{1-{{x}^{2}}} \right)}{\left( 1-2{{x}^{2}} \right)} \right]$ at $x=0$ . For answering this question, we need to assume $x=\tan \theta $ and simplify the function ${{\tan }^{-1}}\left[ \dfrac{\left( \sqrt{1+{{x}^{2}}}-1 \right)}{x} \right]$ and then assume $x=\sin \theta $ and simplify the function ${{\tan }^{-1}}\left[ \dfrac{\left( 2x\sqrt{1-{{x}^{2}}} \right)}{\left( 1-2{{x}^{2}} \right)} \right]$ .

Complete step by step solution:
Now considering from the question, we have been asked to find the derivative of ${{\tan }^{-1}}\left[ \dfrac{\left( \sqrt{1+{{x}^{2}}}-1 \right)}{x} \right]$ with respect to ${{\tan }^{-1}}\left[ \dfrac{\left( 2x\sqrt{1-{{x}^{2}}} \right)}{\left( 1-2{{x}^{2}} \right)} \right]$ at $x=0$ .
Let us assume $x=\tan \theta $ and $u={{\tan }^{-1}}\left[ \dfrac{\left( \sqrt{1+{{x}^{2}}}-1 \right)}{x} \right]$ .
Now by considering these assumptions, we will have
$\begin{align}
  & u={{\tan }^{-1}}\left[ \dfrac{\left( \sec \theta -1 \right)}{\tan \theta } \right] \\
 & \Rightarrow u={{\tan }^{-1}}\left[ \dfrac{1-\cos \theta }{\sin \theta } \right] \\
\end{align}$ .
We know that $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $ and $\csc \theta -\cot \theta =\tan \dfrac{\theta }{2}$ .
By using this we will have
$\begin{align}
  & u={{\tan }^{-1}}\left( \tan \dfrac{\theta }{2} \right) \\
 & \Rightarrow u=\dfrac{\theta }{2} \\
 & \Rightarrow u=\dfrac{1}{2}{{\tan }^{-1}}x \\
\end{align}$ .
Now let us assume $x=\sin {{\theta }_{1}}$ and $v={{\tan }^{-1}}\left[ \dfrac{\left( 2x\sqrt{1-{{x}^{2}}} \right)}{\left( 1-2{{x}^{2}} \right)} \right]$ . By considering these assumptions, we will have
$\begin{align}
  & v={{\tan }^{-1}}\left[ \dfrac{\left( \sin 2{{\theta }_{1}} \right)}{\cos 2{{\theta }_{1}}} \right] \\
 & \Rightarrow v={{\tan }^{-1}}\left( \tan 2{{\theta }_{1}} \right) \\
\end{align}$ .
Since $\sin 2\theta =2\sin \theta \cos \theta $ and $\cos 2\theta =1-2{{\sin }^{2}}\theta $
By simplifying this further we will have $v=2{{\sin }^{-1}}x$ .
Now we need to find the value of$\dfrac{du}{dv}$ for that we will use the chain rule given as $\dfrac{du}{dv}=\dfrac{du}{dx}\times \dfrac{dx}{dv}$ and the formulae given as $\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}$ and $\dfrac{d}{dx}\left( {{\sin }^{-1}}x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ .
By using these things we will have $\dfrac{du}{dv}=\dfrac{1}{2}\left( \dfrac{d}{dx}{{\tan }^{-1}}x \right)\dfrac{dx}{d\left( 2{{\sin }^{-1}}x \right)}$ .
By simplifying this further we will have
$\begin{align}
  & \Rightarrow \dfrac{du}{dv}=\dfrac{1}{4\left( 1+{{x}^{2}} \right)}\left( \dfrac{1}{\left( \dfrac{d}{dx}{{\sin }^{-1}}x \right)} \right) \\
 & \Rightarrow \dfrac{du}{dv}=\dfrac{\sqrt{1-{{x}^{2}}}}{4\left( 1+{{x}^{2}} \right)} \\
\end{align}$ .
At $x=0$ we will have $\dfrac{du}{dv}=\dfrac{1}{4}$ .
Therefore we can conclude that the value of the derivative of ${{\tan }^{-1}}\left[ \dfrac{\left( \sqrt{1+{{x}^{2}}}-1 \right)}{x} \right]$ with respect to ${{\tan }^{-1}}\left[ \dfrac{\left( 2x\sqrt{1-{{x}^{2}}} \right)}{\left( 1-2{{x}^{2}} \right)} \right]$ at $x=0$ is given as $\dfrac{1}{4}$ .
Hence we will mark the option “2” as correct.

Note: While answering questions of this type, we should be sure with the concepts that we are going to apply in between the process. Here we should be going step by step, this question looks to be complex one but by solving step by step it becomes simple.