The derivative of $\sec (\tan \sqrt{x})$ with respect to x is :
Answer & explanation
Correct answer: option 1
$y = \sec (\tan \sqrt{x})$
differentiating wrt x
$\frac{d y}{d x} =\sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \frac{d}{d x}(\tan \sqrt{x})$
$=\frac{\sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \sec ^2 \sqrt{x} \frac{d}{d x} \sqrt{x}}{\frac{d y}{d x}}$ [Using chain rule]
$\frac{d y}{d x}=\frac{\sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \sec ^2 \sqrt{x}}{2 \sqrt{x}}$