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Math Help - derivatives

  1. #1
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    derivatives

    can somebody please help me find the second derivative of y = x squared times e to the 1/x squared

    thanks
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  2. #2
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    Hello, turtle!

    This problem is a killer . . .


    Find the second derivative of: . y \:= \:x^2\!\cdot\!e^{1/x^2}

    We have: . y \;= \;x^2\cdot e^{x^{-2}}


    Product Rule: . y'\;=\;x^2\cdot e^{x^{-2}}(-2x^{-3}) + 2x\cdot e^{x^{-2}} \;=\;-2x^{-1}e^{x^{-2}} + 2xe^{x^{-2}}

    . . . . Factor: . y'\;=\;2e^{x^{-2}}\left(x - x^{-1}\right)


    Product Rule: . y''\;= \;2e^{x^{-2}}\left(1 + x^{-2}\right) + 2e^{x^{-2}}\left(-2x^{-3}\right)\left(x - x^{-1}\right)

    . . . . . . . . . . y''\;=\;2e^{x^{-2}}\left(1 + x^{-2}\right) - 4x^{-3}e^{x^{-2}}\left(x - x^{-1}\right)

    Factor: . y'' \;= \;2e^{x^{-2}}\left[1 + x^{-2} - 2x^{-3}\left(x - x^{-1}\right)\right]

    . . . . . . y'' \;=\;2e^{x^{-2}}\left[1 + x^{-2} - 2x^{-2} + 2x^{-4}\right]

    . . . . . . y''\;=\;2e^{x^{-2}}\left[1 - x^{-2} + 2x^{-4}\right]

    . . . . . . y''\;=\;2e^{x^{-2}}\left[1 - \frac{1}{x^2} + \frac{2}{x^4}\right]

    . . . . . . y''\;=\;2e^{x^{-2}}\left[\frac{x^4 - x^2 + 2}{x^4}\right]

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  3. #3
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    You can also try a different approach that is sometimes helpful, but maybe not in this case which is pretty complicated.

    Start with y = x^2 e^{1/x^2}.

    Take the natural log of both sides:

    \ln y = \ln \left( x^2 e^{1/x^2} \right)

    Expand the logs on the right side:

    \ln y = 2 \ln x + x^{-2}

    Take the derivative (noting the chain rule):

    \frac{1}{y}\,y' = 2x^{-1} - 2x^{-3}

    Move the y over:

    y' = y \left( 2x^{-1} - 2x^{-3} \right)

    Now take the second derivative, applying the product rule on the right side:

    y'' = y' \left( 2x^{-1} - 2x^{-3} \right) + y \left( -2x^{-2} + 6x^{-4} \right)

    Substitute in the y' which was found one step earlier:

    y'' = y \left( 2x^{-1} - 2x^{-3} \right)^2 + y \left( -2x^{-2} + 6x^{-4} \right)

    Factor out the y and simplify the garbage that's left over:

    y'' = y \left[ 2x^{-2} - 2x^{-4} + 4x^{-6} \right]

    Substitute in y (which was the original equation):

    y'' = x^2 e^{1/x^2} \left[ 2x^{-2} - 2x^{-4} + 4x^{-6} \right]

    Oh look, that x^2 can get passed in for a little simplification.
    Also, a 2 can get pulled out:

    y'' = 2e^{1/x^2} \left( 1 - x^{-2} + 2x^{-4} \right)

    In hindsight, this seems to be no less complicated than the first solution provided. Sometimes this technique can save considerable time though, and other times it can over complicate. I like this one though because you can see how with an exponential, the derivative is closely related to the function itself (I enjoy differential equations).
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  4. #4
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    I like it, Soltras!

    I considered log differentiation but decided to do it head-on.

    I especially like the part where \ln\left(e^{^{\frac{1}{x^2}}}\right) becomes x^{-2} . . . delightful!

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