Question

Find dy dx of y equals the secant of the square root of x.

Answer 1

$\bf y=sec(\sqrt{x})\implies y=sec\left( x^{\frac{1}{2}} \right)\implies \cfrac{dy}{dx}=\stackrel{\textit{chain rule}}{sec\left( x^{\frac{1}{2}} \right)tan\left( x^{\frac{1}{2}} \right)\cdot \cfrac{1}{2}x^{-\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=sec\left( x^{\frac{1}{2}} \right)tan\left( x^{\frac{1}{2}} \right)\cdot \cfrac{1}{2x^{\frac{1}{2}}}\implies \cfrac{dy}{dx}=\cfrac{sec\left( \sqrt{x} \right)tan\left( \sqrt{x} \right)}{2\sqrt{x}}$