Let $f(x)= \sin^{-1}(2e^x)$. What is $f '(\ln(3/10)$? So I tried switching $y= \sin^{-1}(2e^x)$ with $x= \sin(2e^y)$ and then finding the derivative. The answer does not seem to match up. Thanks for the help

You have $\sin y=2e^x$, $y'\cos y=2e^x$ and you have $y'=\frac{2e^x}{\cos \sin^{-1}(2e^x)}$.
