Question

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

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

Answer 1

The correct option is D

none of these

Explanation for correct option:

Step-1: Simplify given data.

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

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

Step-2: Differentiate w.r.t $x$

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

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