Continuity and Differentiability — Class 12 Maths Solution

Let$y = \sec \{ \tan (\sqrt x )\}$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{d}{{dx}}\sec (\tan \sqrt x ) = \sec (\tan \sqrt x )\tan (\tan \sqrt x )\cfrac{d}{{dx}}\tan \sqrt x$

$= \sec (\tan \sqrt x ) \cdot \tan (\tan \sqrt x ) \cdot {\sec ^2}\sqrt x \cfrac{d}{{dx}}(\sqrt x )$

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

$= \sec (\tan \sqrt x ) \cdot \tan (\tan \sqrt x ) \cdot {\sec ^2}\sqrt x \cdot \cfrac{1}{{2\sqrt x }}$

therefore $\frac{dy}{dx}=\sec (\tan \sqrt x ) \cdot \tan (\tan \sqrt x ) \cdot {\sec ^2}\sqrt x \cdot \cfrac{1}{{2\sqrt x }}$