I'm not sure if I did the problem right. Any help verifying would be great.
finding the derivative
$y= \arcsin(e^x)$
$\frac{dy}{dx}= \frac{1}{\sqrt{1-(e^x)^2}} \cdot e^x \cdot 1$
I'm not sure if I did the problem right. Any help verifying would be great.
finding the derivative
$y= \arcsin(e^x)$
$\frac{dy}{dx}= \frac{1}{\sqrt{1-(e^x)^2}} \cdot e^x \cdot 1$
Your work looks correct.
$\frac{d}{dx} \arcsin(e^x) = \frac{e^x}{\sqrt{1 - e^{2x}}}$
This site has a good explanation of (and exercises) for derivatives of inverse trigonometric functions, including a derivation of the formula $\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 - x^2}}$.