# Thread: derivative of exponential function

1. ## derivative of exponential function

$\displaystyle f(x)=e^{\frac{-x^2}{2}}$, find $\displaystyle f'(x)$

My textbook says the answer is:

$\displaystyle f'(x) = e^{\frac{-x^2}{2}}(-\frac{2x}{2}) = -xe^{\frac{-x^2}{2}}$

However, there isn't an explanation of the first step. Where does $\displaystyle (-\frac{2x}{2})$ come from? Could someone explain this to me?

Thanks,

2. Originally Posted by absvalue

$\displaystyle f(x)=e^{\frac{-x^2}{2}}$, find $\displaystyle f'(x)$

My textbook says the answer is:

$\displaystyle f'(x) = e^{\frac{-x^2}{2}}(-\frac{2x}{2}) = -xe^{\frac{-x^2}{2}}$

However, there isn't an explanation of the first step. Where does $\displaystyle (-\frac{2x}{2})$ come from? Could someone explain this to me?

Thanks,
It's from the chain rule:
The derivative of f[g(x)] is $\displaystyle f'[g(x)]\cdot g'(x)$

You know the derivative of $\displaystyle e^x$ is $\displaystyle e^x$, right?

So the derivative of $\displaystyle e^{f(x)}$ is $\displaystyle e^{f(x)}\cdot f'(x)$

Hope that helps

3. Oh, I see... that's cool. Thanks!

4. Originally Posted by absvalue
Oh, I see... that's cool. Thanks!
No problem!

By the way, don't forget to do this multiple times, if the function embedded inside is not just x.

I mean the derivative of $\displaystyle f[g(h(x))]$ is:

$\displaystyle f'[g(h(x))]\cdot g'[h(x)] \cdot h'(x)$

Anytime you see a function where the inside is not just x, don't forget the chain rule!

Hope this helps!