Differential coefficient of see ((tan^{-1} x)) with respect... | Mathematics | SolveeIt | Solveeit

Today, August 18

Question

Differential coefficient of see ( $tan^{-1} x$ ) with respect to x is:

A

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

B

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

C

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

D

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

Answer

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

Explanation

Solution

The correct option is (C): $\frac{x}{\sqrt{1 + x^2}}$