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Math Help - Small Question on Integration

  1. #1
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    Small Question on Integration

    Hello, and thanks in advance for the help.

    I know that the derivative of e^x is just e^x. However, if I'm given a function such as e^-x, how is the -x reflected in integrating?

    Something tells me it can't be as simple as e^-x ---> e^-x.

    Thanks again!
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  2. #2
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    Dear NBrunk,

    Here you must use the chain rule.

    \frac{d(e^{-x})}{dx}=\frac{d(e^{-x})}{d(-x)}\times\frac{d(-x)}{dx}

    \frac{d(e^{-x})}{dx}=-e^{-x}

    Hope this helps.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by NBrunk View Post
    Hello, and thanks in advance for the help.

    I know that the derivative of e^x is just e^x. However, if I'm given a function such as e^-x, how is the -x reflected in integrating?

    Something tells me it can't be as simple as e^-x ---> e^-x.

    Thanks again!
    Well, almost that simple....

    Let's make use of the chain rule here, namely that \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}

    So, let y=e^u and let u=-x.

    Then the derivative of e^u would be \frac{dy}{dx}=e^u\cdot\frac{du}{dx}=e^{-x}\cdot(-1)=-e^{-x}
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  4. #4
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     \int e^{-x}~dx = -e^{-x}+C

    You can differentiate  -e^{-x}+C to check
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