# Derivative of $\sin(e^{-x})$

I'm am currently reviewing for a test and I'm not sure what the derivative of $\sin(e^{-x})$ is.

The answer I got is $-e^{-x\cos(e^{-x})}$. Is this correct? If not what is?

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$\left(\sin(u)\right)'=u'\cos(u),~~u=f(x)$ – Babak S. Apr 19 '13 at 15:59

You need to make sure you're using the chain rule properly. The rule is $$f(g(x))' =g'(x) \times f'(g(x))$$ Be careful with the brackets! Can you see where you went wrong?

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It's hard to tell from your formatting, but if you mean $-e^{-x}\cos(e^{-x})$, you're correct.

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No, it is definitely not just some power of $e$. (Original post had: "-e^(-xcos(e^-x)) ")

Apply the chain rule to the original expression.

If $f(x)=\sin(x)$ and $g(x)=e^{-x}$, you are trying to take the derivative of $f(g(x))$. The chain rule says that the derivative should be $f'(g(x))\cdot g'(x)$.

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Here's the Chain Rule:$$\dfrac{d}{dx}f(g(x)) = f'(g(x))g'(x)$$Let $f(x) = \sin(x)$ and $g(x) = e^{-x}$.

Substituting the proper values, we get:$$\cos(e^{-x})\times -e^{-x} = -e^{-x}\cos(e^{-x})$$

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You need to apply the chain rule. Let $u = \operatorname{e}^{-x}$ so that $\sin(\operatorname{e}^{-x}) = \sin u$. The chain rule says:

\begin{array}{ccc} \frac{\operatorname{d}\!f}{\operatorname{d}\!x} &=& \frac{\operatorname{d}\!f}{\operatorname{d}\!u} \times \frac{\operatorname{d}\!u}{\operatorname{d}\!x} \\ \\ \\ &=& (\cos u) \times (-\operatorname{e}^{-x}) \\ \\ &=&-\operatorname{e}^{-x}\cos(\operatorname{e}^{-x}) \end{array}

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