Question

$\frac{d\left({\tan }^{-1}\left({\displaystyle \frac{a-x}{1+ax}}\right)\right)}{dx}=$

• A. $\frac{1}{\left(1+x^2\right)}$
• B. $-\frac{1}{\left(1+x^2\right)}$
• C. $\frac{a}{\left(1+x^2\right)}$
• D. none of these

Answer 1

The correct option is B

$-\frac{1}{\left(1+x^2\right)}$

Explanation for correct option:

Step-1: Simplify the given data.

Given, $\frac{d\left({\tan }^{-1}\left({\displaystyle \frac{a-x}{1+ax}}\right)\right)}{dx}$

$=\frac{d\left({\tan }^{-1}\left({\displaystyle a}\right)-{\tan }^{-1}\left(x\right)\right)}{dx}\left(\because {\tan }^{-1}\left(\frac{a-x}{1+ax}\right){\tan }^{-1}\left(a\right)-{\tan }^{-1}\left(x\right)\right)$

Step-2: Differentiate with respect to $x$.

$$\begin{gathered} =0-\frac{1}{\left(1+x^2\right)}\left(\because {\tan }^{-1}\left(a\right)\mathrm{is}\mathrm{constant}\right) \\ =-\frac{1}{\left(1+x^2\right)} \end{gathered}$$

Hence, correct answer is option $\left(B\right)$