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Math Help - Second Derivative Help

  1. #1
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    Second Derivative Help

    I am stuck on a problem concerning second derivatives and was wondering if you could provide
    some help.

    orignial function:

    f(x)= (e^x)/x


    I believe i figured out the first derivative using the quotient rule.

    f'(x)= [e^x(x-1)] / x^2


    Now, I am not sure what to do with the top. I know the chain rule is
    somewhere in there. Any advice would be great, thank you.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lostsheep View Post
    I am stuck on a problem concerning second derivatives and was wondering if you could provide
    some help.

    orignial function:

    f(x)= (e^x)/x


    I believe i figured out the first derivative using the quotient rule.

    f'(x)= [e^x(x-1)] / x^2


    Now, I am not sure what to do with the top. I know the chain rule is
    somewhere in there. Any advice would be great, thank you.
    the product rule actually:

    f'(x) = \frac {e^x ( x - 1)}{x^2}

    By the quotient rule, we get:

    f''(x) = \frac { \left( x^2 \right)( e^x (x - 1))' - (e^x(x - 1)) \left( x^2 \right)'}{x^4}

    = \frac {x^2[e^x(x - 1) + e^x] - 2x(e^x(x - 1))}{x^4}

    and simplify

    and that's using the quotient rule. were it me though, i'd change the function so i can use the product rule all the way, i find the quotient rule very annoying
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  3. #3
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    I think you may separate it into two parts, as

    f''(x)=x(e^x)/(x^2) d/dx - (e^x)/(x^2) d/dx
    =(e^x)/x d/dx - (e^x)/(x^2) d/dx
    =(e^x(x-1))/(x^2) - ((x^2*e^x)-(e^x(4*x^3))/(x^4)
    ...

    and simplify it...

    I hope it is right..
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    the product rule actually:

    f'(x) = \frac {e^x ( x - 1)}{x^2}

    By the quotient rule, we get:

    f''(x) = \frac { \left( x^2 \right)( e^x (x - 1))' - (e^x(x - 1)) \left( x^2 \right)'}{x^4}

    = \frac {x^2[e^x(x - 1) + e^x] - 2x(e^x(x - 1))}{x^4}

    and simplify

    and that's using the quotient rule. were it me though, i'd change the function so i can use the product rule all the way, i find the quotient rule very annoying

    How would you change the function to use only the product rule?

    As for simplifying, I got

    [e^x(x-1+1)(-2x^2 -2x) ] / x^2


    Is that even remotely close? Thanks for all the help.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lostsheep View Post
    How would you change the function to use only the product rule?
    like this:

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

    \Rightarrow f(x) = x^{-1}e^x

    \Rightarrow f'(x) = -x^{-2}e^x + x^{-1}e^x

    \Rightarrow f''(x) = 2x^{-3}e^x + -x^{-2}e^x - x^{-2}e^x + x^{-1}e^x

    = e^x \left( 2x^{-3} - 2x^{-2} + x^{-1} \right)

    and simplify that. easy huh?



    As for simplifying, I got

    [e^x(x-1+1)(-2x^2 -2x) ] / x^2


    Is that even remotely close? Thanks for all the help.
    that is not simplified. for one obvious reason, we could write x - 1 + 1 as just x. there are other ways you can simplify
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    like this:

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

    \Rightarrow f(x) = x^{-1}e^x

    \Rightarrow f'(x) = -x^{-2}e^x + x^{-1}e^x

    \Rightarrow f''(x) = 2x^{-3}e^x + -x^{-2}e^x - x^{-2}e^x + x^{-1}e^x

    = e^x \left( 2x^{-3} - 2x^{-2} + x^{-1} \right)

    and simplify that. easy huh?


    that is not simplified. for one obvious reason, we could write x - 1 + 1 as just x. there are other ways you can simplify

    That is very interesting, i suppose you can just elminate the quotiet rule altogether.

    Simplifing is one of my weak points.

    so thats say I did use the product rule.

    f''(x)= e^x (2x^-3 -2x^-2 + x^-1)

    then simplified would be;

    f''(x) = xe^x ( 2x^-2 -2x^-1 -1)

    f''(x) = xe^x / (2x^2 -2x -1 )


    If that is not corrected, could you provide it simplified in using fraction/ division. I want to realize my mistakes. Thanks a lot.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lostsheep View Post
    That is very interesting, i suppose you can just elminate the quotiet rule altogether.

    Simplifing is one of my weak points.

    so thats say I did use the product rule.

    f''(x)= e^x (2x^-3 -2x^-2 + x^-1)

    then simplified would be;

    f''(x) = xe^x ( 2x^-2 -2x^-1 -1)

    f''(x) = xe^x / (2x^2 -2x -1 )


    If that is not corrected, could you provide it simplified in using fraction/ division. I want to realize my mistakes. Thanks a lot.
    e^x \left( 2x^{-3} - 2x^{-2} + x^{-1} \right)

    = e^x \left( \frac{2}{x^3} - \frac {2}{x^2} + \frac {1}{x} \right)

    = e^x \left( \frac{x^2 - 2x + 2}{x^3} \right)


    you factored out an x incorrectly, so you were wrong from the beginning. you could have factored out the x^{-3} and you would have gotten what i did almost immediately. try it
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    e^x \left( 2x^{-3} - 2x^{-2} + x^{-1} \right)

    = e^x \left( \frac{2}{x^3} - \frac {2}{x^2} + \frac {1}{x} \right)

    = e^x \left( \frac{x^2 - 2x + 2}{x^3} \right)


    you factored out an x incorrectly, so you were wrong from the beginning. you could have factored out the x^{-3} and you would have gotten what i did almost immediately. try it
    Thank you very much. I need to work on that. Natural Logs and e just throw me off when it comes to this, as well as the simplifing. Thanks again.
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