Question

The derivative of $\ln (1+x^2)$ with respect to ${\tan }^{-1}x$ is

[1 mark]

• A. $2x$
• B. $x^2$
• C. $x$
• D. $x^3$

Answer 1

The correct option is A $2x$

Let $y=\ln (1+x^2),z={\tan }^{-1}\left(x\right)$
Now, $\frac{dy}{dx}=\frac{2x}{1+x^2}$
Also, $\frac{dz}{dx}=\frac{1}{1+x^2}$
$\therefore \frac{dy}{dz}=\frac{\frac{dy}{dx}}{\frac{dz}{dx}}=\frac{\frac{2x}{1+x^2}}{\frac{1}{1+x^2}}=2x$