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Math Help - Find the derivative of this product

  1. #1
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    Find the derivative of this product

    \sqrt{z} * e^{-z}
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by sk8erboyla2004 View Post
    \sqrt{z} * e^{-z}
    \sqrt{z} * e^{-z}

    *Chain rule & product rule

    First rewrite as:

    (z)^{\frac{1}{2}} \cdot e^{-z}

    Take a stab at it and come back if you're still stuck!
    Last edited by mr fantastic; March 13th 2009 at 10:39 PM.
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    \sqrt{z} * e^{-z}

    Find derivative

    This is the answer I came up with

    \frac{1}{2\sqrt{z}}e^{-z}+\sqrt{z}e^{-z}

    however the solution the manual was rather subtraction instead of addition ?

    Also

    \sqrt{(x^{2}*5^{x})^{3}}

    Urgent response needed
    find the derivative

    I got this not really simplified and im not sure how to

    \frac{3}{2}\sqrt{(x^{2}*5^{x})}[5^{x}(2x+x^{2}(ln5)]
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  4. #4
    MHF Contributor matheagle's Avatar
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    Almost, note my correction....

    Quote Originally Posted by sk8erboyla2004 View Post
    \sqrt{z} * e^{-z}

    Find derivative

    This is the answer I came up with

    \frac{1}{2\sqrt{z}}e^{-z}-\sqrt{z}e^{-z}

    however the solution the manual was rather subtraction instead of addition ?

    Also

    \sqrt{(x^{2}*5^{x})^{3}}

    Urgent response needed
    find the derivative

    I got this not really simplified and im not sure how to

    \frac{3}{2}\sqrt{(x^{2}*5^{x})}[5^{x}(2x+x^{2}(ln5)]
    Because the derivative of e^{-z} is e^{-z} times the derivative of it's exponent
    which is -z.

    As for the second one I would simplify first. Note that

    (x^2\cdot 5^x)^{3/2}=  x^3\cdot 5^{3x/2}, which is easier to work with.
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  5. #5
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by sk8erboyla2004 View Post
    \sqrt{z} * e^{-z}

    find derivative

    this is the answer i came up with

    \frac{1}{2\sqrt{z}}e^{-z}+\sqrt{z}e^{-z}

    however the solution the manual was rather subtraction instead of addition ? the reason the book answer was subtraction, is because they moved the e^{-z} to the denominator and turned it into \frac{\sqrt{z}}{e^z} & used the quotient rule.
    so you would have:
    \frac{\sqrt{z}}{e^z}

    = \frac{z^{\frac{1}{2}}e^z - \frac{1}{2}z^{-\frac{1}{2}}e^z}{e^{2z}}


    also

    \sqrt{(x^{2}*5^{x})^{3}}

    urgent response needed
    find the derivative

    i got this not really simplified and im not sure how to

    \frac{3}{2}\sqrt{(x^{2}*5^{x})}[5^{x}(2x+x^{2}(ln5)]

    see above in red! For the second one, change the square root into ^{\frac{1}{2}} and then do your chain rule. Then when you are finished taking the derivative, you can change any of those left back into roots.
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  6. #6
    MHF Contributor matheagle's Avatar
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    I would not use the chain rule on the second and there's no reason to do a quotient rule on the first.
    He/she left out the derivative of -z.
    Last edited by matheagle; March 13th 2009 at 10:34 PM.
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  7. #7
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    Quote Originally Posted by sk8erboyla2004 View Post
    \sqrt{z} * e^{-z}

    Find derivative

    This is the answer I came up with

    \frac{1}{2\sqrt{z}}e^{-z}+\sqrt{z}e^{-z}

    however the solution the manual was rather subtraction instead of addition ?
    Mr F says: That's because the derivative of {\color{red}e^{-z}} is {\color{red}- e^{-z}}.

    [snip]
    ..
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  8. #8
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by matheagle View Post
    I would not use the chain rule on the second and there's no reason to do a quotient rule on the first.
    He/she left out the derivative of -z.
    How would you do the second without the chain rule?

    \sqrt{(x^{2}*5^{x})^{3}}
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  9. #9
    MHF Contributor matheagle's Avatar
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    Using (ab)^c=a^cb^c we can simplify this to

    x^3(5^{3/2})^x and I would just use the product rule...

    and get x^3(5^{3/2})^x\ln(5^{3/2})+3x^2(5^{3/2})^x.

    This is esier to simplifier than the one you get via the chain rule...

    x^2(5^{3/2})^x\biggl( x\ln(5^{3/2})+3\biggr).
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  10. #10
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by matheagle View Post
    Using (ab)^c=a^cb^c we can simplify this to

    x^3(5^{3/2})^x and I would just use the product rule...

    and get x^3(5^{3/2})^x\ln(5^{3/2})+3x^2(5^{3/2})^x.

    This is esier to simplifier than the one you get via the chain rule.
    Ah, yes. I like making more work for myself
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  11. #11
    MHF Contributor matheagle's Avatar
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    Not many people would notice that.
    I bet the solution manual used the chain rule.
    You're way too hard on yourself.
    I like solving problems several ways, that way I can check my answers.
    Last edited by matheagle; March 14th 2009 at 02:11 PM.
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  12. #12
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    Quote Originally Posted by mr fantastic View Post
    ..

    care to further elaborate how it is negative?
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  13. #13
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by sk8erboyla2004 View Post
    care to further elaborate how it is negative?
    The derivative of e^{g(x)} is g'(x)e^{g(x)}.

    So the derivative of e^{-x} is (-1)e^{-x}=-e^{-x}.
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  14. #14
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    Quote Originally Posted by matheagle View Post
    The derivative of e^{g(x)} is g'(x)e^{g(x)}.
    Or, more specifically generally, the derivative of e^{kz} is k e^{kz}.
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  15. #15
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    It seem to me that


    f(x)=e^{x}


    f\prime(x)=e^{x}

    When does that rule come into play?
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