Question

What is the derivative of ${\tan }^{-1}\left\{\frac{\left(1+x\right)}{\left(1-x\right)}\right\}$ with respect to $x$ ?

• A. $1+x^2$
• B. $\frac{1}{1+x^2}$
• C. $-\frac{1}{1+x^2}$
• D. None of these

Answer 1

The correct option is B

$\frac{1}{1+x^2}$

Explanation of correct answer :

Finding derivative of ${\tan }^{-1}\left\{\frac{\left(1+x\right)}{\left(1-x\right)}\right\}$ with respect to $x$:

Let, $y=$${\tan }^{-1}\left\{\frac{\left(1+x\right)}{\left(1-x\right)}\right\}$

Put, $x=\tan \varnothing$

$\Rightarrow \varnothing ={\tan }^{-1}\left(x\right)$

Now solving $y$.

$\Rightarrow y={\tan }^{-1}\left\{\frac{\left(1+\tan \varnothing \right)}{\left(1-\tan \varnothing \right)}\right\}$

We know that, $\tan \frac{\mathrm{\pi }}{4}=1$,

$$\begin{gathered} \Rightarrow y={\tan }^{-1}\left\{\frac{\left(\tan {\displaystyle \frac{\mathrm{\pi }}{4}}+\tan \varnothing \right)}{\left(1-\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\times \tan \varnothing \right)}\right\}(\tan \varnothing \text{can be written as}1\times \tan \varnothing ) \\ ={\tan }^{-1}\left\{\tan \left(\frac{\mathrm{\pi }}{4}+\varnothing \right)\right\}\left(\mathrm{since}\tan \left(x+y\right)=\frac{\tan x+\tan y}{1-\tan x\tan y}\right) \\ =\frac{\mathrm{\pi }}{4}+\varnothing \\ =\frac{\mathrm{\pi }}{4}+{\tan }^{-1}x \end{gathered}$$

Differentiating $y$ with respect to $x$.

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\frac{d\left({\displaystyle \frac{\mathrm{\pi }}{4}}+{\tan }^{-1}x\right)}{dx} \\ =\frac{1}{1+x^2} \end{gathered}$$

Hence, the derivative of ${\tan }^{-1}\left\{\frac{\left(1+x\right)}{\left(1-x\right)}\right\}$ with respect to $x$ is $\frac{1}{1+x^2}$.

Thus, the correct answer is Option(B).