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Math Help - Second Derivative Value via Implicit Differentiation

  1. #1
    Super Member 11rdc11's Avatar
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    Second Derivative Value via Implicit Differentiation

    Hi guys, I haven't posted in a while but I was tutoring someone and ran into a problem that I thought had got right but blackboard said was incorrect.

    The problem is

     \frac{y^{3}}{3} +2y -x^2 + \frac{5}{3} = 0

    Now find the second derivative when x = 2 and y =1

    I keep getting 3/4 as my answer. I solved it using implicit differentiation and also using partials differentiation to double check. The correct answer according to blackboard is -14/27

    and can someone show me what im doing wrong with my latex. Guess im a bit rusty. Thanks
    Last edited by CaptainBlack; August 3rd 2011 at 11:58 PM. Reason: change math to tex tags
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  2. #2
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    re: Second Derivative Value via Implicit Differentiation

    Quote Originally Posted by 11rdc11 View Post

    and can someone show me what im doing wrong with my latex. Guess im a bit rusty. Thanks
    Use "tex" tags, "math" tags are no good here at the moment

    \displaystyle \frac{y^3}{3}+2y-x^2 + \frac{5}{3} = 0
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    re: Second Derivative Value via Implicit Differentiation

    Quote Originally Posted by 11rdc11 View Post
    Hi guys, I haven't posted in a while but I was tutoring someone and ran into a problem that I thought had got right but blackboard said was incorrect.

    The problem is

     \frac{y^{3}}{3} +2y -x^2 + \frac{5}{3} = 0

    Now find the second derivative when x = 2 and y =1

    I keep getting 3/4 as my answer. I solved it using implicit differentiation and also using partials differentiation to double check. The correct answer according to blackboard is -14/27

    and can someone show me what im doing wrong with my latex. Guess im a bit rusty. Thanks
    Your function is defined by \frac{y^{3}}{3} +2y -x^2 + \frac{5}{3} = 0 . Deriving: y^2y'+2y'-2x=0 . Substituting x=2,y=1 we get 3y'-4=0 or equivalently y'=4/3 . Derive again and substitute x=2,y=1,y'=4/3 .
    Last edited by FernandoRevilla; August 3rd 2011 at 10:45 PM.
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  4. #4
    Super Member 11rdc11's Avatar
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    re: Second Derivative Value via Implicit Differentiation

    Thanks guys but even when i take the second derivative i get the same

    y' = \frac{2x}{y^2+2}

    now taking the derivative again

    y'' = \frac{\frac{2}{y^2+2}}{\frac{4xy}{(y^2+2)^2}}

    which simplifies to

    y'' = \frac{y^2+2}{2xy}

    and plugging in the values you 3/4 right?
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  5. #5
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    re: Second Derivative Value via Implicit Differentiation

    Quote Originally Posted by 11rdc11 View Post
    y'' = \frac{y^2+2}{2xy}

    and plugging in the values you 3/4 right?
    Looks good to me.
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  6. #6
    Super Member 11rdc11's Avatar
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    re: Second Derivative Value via Implicit Differentiation

    Thanks that is what I thought but it kept telling me that I was wrong.
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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    re: Second Derivative Value via Implicit Differentiation

    Deriving y^2y'+2y'-2x=0 we get 2y(y')^2+y^2y''+2y''-2=0 . Substituting x=2,y=1,y'=4/3 you'll obtain y''(2)=-14/27 .
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  8. #8
    Grand Panjandrum
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    re: Second Derivative Value via Implicit Differentiation

    Quote Originally Posted by pickslides View Post
    Use "tex" tags, "math" tags are no good here at the moment

    \displaystyle \frac{y^3}{3}+2y-x^2 + \frac{5}{3} = 0
    You mean "Jedi math tags do not work on MHF ony tex..."

    CB
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  9. #9
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    Re: Second Derivative Value via Implicit Differentiation

    Quote Originally Posted by 11rdc11 View Post
    y' = \frac{2x}{y^2+2}

    now taking the derivative again

    y'' = \frac{\frac{2}{y^2+2}}{\frac{4xy}{(y^2+2)^2}}
    A point against the quotient rule? Many people prefer to see the quotient as a product, anyway. Just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    But this is wrapped inside the legs-uncrossed version of...



    ... the product rule, where, again, straight continuous lines are differentiating downwards with respect to x.




    FR's method is another way of avoiding the quotient rule here. In balloon sculpture...




    __________________________________________________ __________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
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