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Math Help - How would I derive this?

  1. #1
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    How would I derive this?

    y=x^2(e^{-2x})

    What steps would I have to do to get the derivative?
    Last edited by mr fantastic; June 5th 2010 at 03:17 PM. Reason: Added latex tags.
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  2. #2
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    Quote Originally Posted by kmjt View Post
    y=x^2(e^{-2x})

    What steps would I have to do to get the derivative?
    Use the product rule. Please show all your work and say where you're stuck if you need more help.
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    Quote Originally Posted by kmjt View Post
    y=x^2(e^{-2x})

    What steps would I have to do to get the derivative?
    Use the product rule (and don't forget the chain rule on e^{-2x})
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    y'= (2x)(e^-2x) + (x^2)(-2e^-2x)

    I haven't derived for so long, how would I simplify
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  5. #5
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    Quote Originally Posted by kmjt View Post
    y'= (2x)(e^-2x) + (x^2)(-2e^-2x)

    I haven't derived for so long, how would I simplify
    Factorise - There is a common factor ....
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    Quote Originally Posted by kmjt View Post
    y=x^2(e^{-2x})

    What steps would I have to do to get the derivative?
    You can also use logs!

     lny = lnx^2 + lne^{-2x}

     lny = 2lnx -2x

    Implicitly derive

     \frac{1}{y} y^{ \prime } = \frac{2}{x} - 2

     y^{ \prime } = ( \frac{2}{x} - 2 )x^2(e^{-2x})
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  7. #7
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    I ended up getting xe^-2x [ 2 - 2x ] but how do I set this equal to 0 to find the intervals of increase and decrease?
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  8. #8
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    Quote Originally Posted by kmjt View Post
    I ended up getting xe^-2x [ 2 - 2x ] but how do I set this equal to 0 to find the intervals of increase and decrease?
    From the null factor law: xe^{-2x} (2 - 2x) = 0 \Rightarrow x = 0 or 2 - 2x = 0. Note that one of the solutions does NOT correspond to a turning point.
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