Question

Assertion ($\text{A}$): Derivative of $\frac{{\tan }^{-1}x}{1+{\tan }^{-1}x}$ with respect to ${\tan }^{-1}x$ is $\frac{1}{(1+{\tan }^{-1}x{)}^2}$
Reason ($\text{R}$): Derivative of $\frac{x}{1+x}$ with respect to $x$ is $\frac{1}{(1+x{)}^2}$

• A. A is correct R is wrong.
• B. A is wrong R is correct.
• C. A is correct R is correct and R is the correct explanation of A.
• D. A is correct R is correct but R is not the correct explanation of A.

Answer 1

Answer: (C)

A is correct R is correct and R is the correct explanation of A.

Assertion:

$\frac{d}{d{\tan }^{-1}x}\left(\frac{{\tan }^{-1}x}{1+{\tan }^{-1}x}\right)$ $=\frac{1\cdot ({\tan }^{-1}x+1)-1\cdot {\tan }^{-1}x}{({\tan }^{-1}x+1{)}^2}$ $=\frac{1}{({\tan }^{-1}x+1{)}^2}$
Reason:

$\frac{d\left(\frac{x}{1+x}\right)}{dx}=\frac{1\cdot (1+x)-x\cdot 1}{(1+x{)}^2}=\frac{1}{(1+x{)}^2}$

Hence, option C.