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Thread: Another calculating y'

  1. #1
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    Another calculating y'

    If y=xe^(-1/x)

    is y'= xe^(-1/x)*(-(x-1)/x^2)+e^(-1/x)
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  2. #2
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    Quote Originally Posted by spazzyskylar View Post
    If y=xe^(-1/x)

    is y'= xe^(-1/x)*(-(x-1)/x^2)+e^(-1/x)

    You said
    d/dx(-1/x) = -(x-1) / x^2

    Are you sure about this?
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  3. #3
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    Quote Originally Posted by spazzyskylar View Post
    If y=xe^(-1/x)

    is y'= xe^(-1/x)*(-(x-1)/x^2)+e^(-1/x)
    ?

    note that the derivative of $\displaystyle -\frac{1}{x}$ is $\displaystyle \frac{1}{x^2}$
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  4. #4
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    Not really, should it be -(x+1)/x^2
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  5. #5
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    Not quite! if $\displaystyle u=-\frac{1}{x}$

    $\displaystyle \frac{d}{dx}(xe^\frac{-1}{x})=\frac{d}{dx}xe^u=x\frac{d}{dx}e^u+e^u(1)$

    $\displaystyle =x\frac{du}{du}\frac{d}{dx}e^u+e^u=x\frac{du}{dx}\ frac{d}{du}e^u+e^u$

    $\displaystyle =(x)\frac{1}{x^2}e^u+e^u=\frac{e^\frac{-1}{x}}{x}+e^\frac{-1}{x}$
    Last edited by Archie Meade; Jan 3rd 2010 at 02:07 PM.
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  6. #6
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    Quote Originally Posted by spazzyskylar View Post
    If y=xe^(-1/x)

    is y'= xe^(-1/x)*(-(x-1)/x^2)+e^(-1/x)
    $\displaystyle u = x \: \rightarrow \: u' = 1$

    $\displaystyle v = e^{-\frac{1}{x}} \: \: \rightarrow \: v' = \frac{1}{x^2}e^{-\frac{1}{x}}$

    skeeter gave the derivative of the exponent so multiply due to the chain rule

    Now use the product rule

    y' = u'v + uv'

    Spoiler:
    $\displaystyle e^{-\frac{1}{x}} + \frac{1}{x}e^{-\frac{1}{x}} = e^{-\frac{1}{x}}\left(1+\frac{1}{x}\right)$
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  7. #7
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    I got it now
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  8. #8
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    $\displaystyle y'=x'e^{-1\over x} +x(e^{-1\over x})'$
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