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

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

The correct answer is Option (3) → $\frac{x}{\sqrt{1+x^2}}$

$y=\sec(\tan^{-1}x)$

so $\frac{dy}{dx}=\frac{\sec(\tan^{-1}x)\tan(\tan^{-1}x)}{(1+x^2)}$

from this

$\sec(\tan^{-1}x)=\sqrt{1+x^2}$

so $\frac{dy}{dx}=\frac{\sqrt{1+x^2}×x}{1+x^2}$

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