Question

√sec(2x+1) derivative of the function​

Answer 1

$y = \sqrt{sec(2x + 1)} \\ \\ \frac{dy}{dx} = \frac{1}{2 \sqrt{sec(2x + 1)} } \frac{d}{dx} sec(2x + 1) \\ \\ \frac{dy}{dx} = \frac{1}{2 \sqrt{sec(2x + 1)} } sec(2x + 1)tan(2x + 1) \frac{d}{dx} (2x + 1) \\ \\ \frac{dy}{dx} = \frac{2sec(2x + 1)tan(2x + 1)}{2 \sqrt{sec(2x + 1)} } \\ \\ \frac{dy}{dx} = \frac{sec(2x + 1)tan(2x + 1)}{ \sqrt{sec(2x + 1)} }$