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Math Help - Another chain rule question.

  1. #1
    Junior Member madmax29's Avatar
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    Another chain rule question.

    Can you also check this out for me;

    y=e^a

    where a=-x^2/2

    1... dy/dx = dy/du \cdot du/dx

    2...where y = e ^-^u and u= - x^2/2

    3.... dy/du=e ^-^u and du/dx= -x <<<< can you explain how step 2 for "u" this turns into -x

    4.... therefore; dy/du \cdot du/dx = -x e ^ - x^2/2
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by madmax29 View Post
    Can you also check this out for me;

    y=e^a

    where a=-x^2/2

    1... dy/dx = dy/du \cdot du/dx

    2...where y = e ^-^u and u= - x^2/2

    3.... dy/du=e ^-^u and du/dx= -x <<<< can you explain how step 2 for "u" this turns into -x

    4.... therefore; dy/du \cdot du/dx = -x e ^ - x^2/2

    You are using a instead of u. Minor difference but instead you need to find

    1. \dfrac{dy}{dx} = \dfrac{dy}{da} \cdot \dfrac{da}{dx}

    2. Aside from my above note about a then yes.

    3. You need to use the chain rule. It's a tricky one to spot but if you imagine e^{-a} = e^{-1 \cdot a} so the derivative is multiplied by -1. Thus \dfrac{d}{da}e^{-a} = -e^{-a}

    For \dfrac{da}{dx} = -x is true because you can write it as -\dfrac{1}{2} \cdot x^2. x^2 is simple enough but you must multiply by -1/2 which gives -x

    4. Almost, because of the sign error in finding \dfrac{da}{dx} your answer is a factor of -1 out.

    \dfrac{dy}{dx} = -x \cdot -e^{-x^2/2} = xe^{-x^2/2}
    Last edited by e^(i*pi); November 7th 2010 at 05:58 AM. Reason: adding [/math] tag
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