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Math Help - derivative of exponential function

  1. #1
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    derivative of exponential function

    I'm confused about this problem:

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

    My textbook says the answer is:

    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 (-\frac{2x}{2}) come from? Could someone explain this to me?

    Thanks,
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  2. #2
    Member mathemagister's Avatar
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    Quote Originally Posted by absvalue View Post
    I'm confused about this problem:

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

    My textbook says the answer is:

    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 (-\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 f'[g(x)]\cdot g'(x)

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

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

    Hope that helps
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  3. #3
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    Oh, I see... that's cool. Thanks!
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  4. #4
    Member mathemagister's Avatar
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    Quote Originally Posted by absvalue View Post
    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 f[g(h(x))] is:

    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!
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