Question

3. Differentiate: y =
$\sqrt{sec} \sqrt{x}$
sec
sec Vx​

Answer 1

Step-by-step explanation:

We have,

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

Differentiating both sides,we get,

$\implies \frac{dy}{dx} = \frac{ \sec( \sqrt{x} ) \tan( \sqrt{x} ) }{2 \sqrt{ \sec( \sqrt{x} ) } } \times \frac{1}{2 \sqrt{x} }$

$\implies \frac{dy}{dx} = \frac{ \sqrt{ \sec( \sqrt{x} ) } \tan( \sqrt{x} ) }{4 \sqrt{x} }$