# Specific steepness of function

How can I find the point of a function where the steepness is exactly 45 degrees?

In my specific case, the function is $\dfrac{-1}{\exp(x)}$.

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Well, $\tan(45^\circ)=\frac{\mathrm d}{\mathrm dx}(-\exp(-x))$ is the equation to solve... –  Ｊ. Ｍ. Aug 29 '11 at 16:00

Let's say our function is $f(x)$. Can you think of what the value of $f'(x)$, the derivative of $f$, is at a point $x$ where "the steepness is exactly 45 degrees"? Drawing a picture might help.
Do you know how to find the derivative of $f(x)=-\dfrac{1}{e^x}$?