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Math Help - Derivative of Exponential Function

  1. #1
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    Derivative of Exponential Function

    I'm still having a bit of trouble with the concept of the derivative of the exponential function (e^x).

    "If h(z) = (z^2)(1+e^-z), determine h'(-1)."

    How do I solve this?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    ok

    Quote Originally Posted by Jeavus View Post
    I'm still having a bit of trouble with the concept of the derivative of the exponential function (e^x).

    "If h(z) = (z^2)(1+e^-z), determine h'(-1)."

    How do I solve this?
    ok you have h(x)=x^2(1+e^{-x})...then applying the product rule we get h'(x)=x^2\cdot{-e^{-x}}+2x(1+e^{-x})
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Maybe

    this will be helpful \frac{D[e^{u(x)}]}{dx}=e^{u(x)}\cdot{u'(x)}...so for e^{-x} we have u(x)=-x,u'(x)=-1....so using the formula I have you we have \frac{D[e^{-x}]}{dx}=e^{-x}\cdot{-1}=-e^{-x} as indicated above
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by Jeavus View Post
    I'm still having a bit of trouble with the concept of the derivative of the exponential function (e^x).

    "If h(z) = (z^2)(1+e^-z), determine h'(-1)."

    How do I solve this?
    h(z)=z^2+z^2e^{-z} taking the derivative

     \frac{dh}{dz}=2z+\underbrace{2ze^{-z}-z^2e^{-z}}_{productRule}

     \frac{dh}{dz}|_{z=-1}=2(-1)+2(-1)e^{1}-(-1)^2e^{1}=-2-3e
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  5. #5
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    Thanks you two!
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Haha

    Quote Originally Posted by Jeavus View Post
    Thanks you two!
    Its not as though I have an english term paper due for English 11 tomorrow...haha xD
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