Question

If $y={\sin }^{-1}\left(2x\right)$ then find $\frac{dy}{dx}$
or
f $y=x^{2x}$ then find $\frac{dy}{dx}$.

Answer 1

$y={\sin }^{-1}\left(2x\right)$

$\frac{dy}{dx}=\frac{2}{\sqrt{1-4x^2}}$

for $y=x^{2x}\log y=2x\log x\frac{1}{y}\frac{dy}{dx}=2(\frac{x}{x}+\log x)=2(1+\log x)$

So, $\frac{dy}{dx}=2y(1+\log x)=2x^{2x}(1+\log x)$