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Thread: Derivative of e.

  1. #1
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    Derivative of e.

    Please check if i have done these correctly.

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

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

    $\displaystyle f''(x)= -4x^2e^{-x^2} - 2e^{-x^2}$


    Can i also write f''(x) as $\displaystyle -4e^{-x^2}x^2-2e^{-x^2}$ ??
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  2. #2
    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by el123 View Post
    Please check if i have done these correctly.

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

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

    $\displaystyle f''(x)= -4x^2e^{-x^2} - 2e^{-x^2}$


    Can i also write f''(x) as $\displaystyle -4e^{-x^2}x^2-2e^{-x^2}$ ??
    Your $\displaystyle f''(x)$ is incorrect as it should be $\displaystyle 4x^2e^{-x^2} - 2e^{-x^2}$ since we started with a factor of $\displaystyle -2x$ and multiplied it by $\displaystyle -2x$ again when we use the chain rule to get $\displaystyle -2x*-2x = 4x^2$. And as for your re-writing of $\displaystyle f''(x)$, you just moved the $\displaystyle x^2$ term around...that's not different in any meaningful way ie. just like $\displaystyle ax^2$ and $\displaystyle x^2a$ are obviously the same since multiplication in $\displaystyle \mathbb{R}$ is commutative. If you wish to factor $\displaystyle f''(x)$ you can rewrite it as $\displaystyle f''(x) = (4x - 2)e^{-x^2}$
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