# finding the derivative of an arcsin

I'm not sure if I did the problem right. Any help verifying would be great.

finding the derivative

$$y= \arcsin(e^x)$$

$$\frac{dy}{dx}= \frac{1}{\sqrt{1-(e^x)^2}} \cdot e^x \cdot 1$$

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Looks good; but you may want another set of parentheses around the square root term to avoid ambiguity. –  David Mitra May 17 '12 at 21:01

$$\frac{d}{dx} \arcsin(e^x) = \frac{e^x}{\sqrt{1 - e^{2x}}}$$

This site has a good explanation of (and exercises) for derivatives of inverse trigonometric functions, including a derivation of the formula $\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 - x^2}}$.

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thank you i will see that site –  dave5678 May 17 '12 at 21:19