# How do you find the derivative of  y=arcsin(e^x)?

Sep 10, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} / \sqrt{1 - {e}^{2 x}}$

#### Explanation:

Alternative approach (basically the same as the one already presented):

Note that $y = \arcsin \left({e}^{x}\right)$ can be manipulated to say that $\sin \left(y\right) = {e}^{x}$.

Take the derivative of both sides with respect to $x$. Recall that the chain rule will be used on the left side: $\cos \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x}$

Solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$ gives $\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} / \cos \left(y\right)$. We know that $\sin \left(y\right) = {e}^{x}$, so we can write $\cos \left(y\right)$ as cos(y)=sqrt(1-sin^2(y). Note that this relationship comes from the Pythagorean identity ${\sin}^{2} \left(y\right) + {\cos}^{2} \left(y\right) = 1$.

Thus, we have $\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} / \sqrt{1 - {\sin}^{2} \left(y\right)}$. And, since we know that $\sin \left(y\right) = {e}^{x}$, this gives a fully simplified derivative of $\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} / \sqrt{1 - {e}^{2 x}}$.