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Implicit differentiation, whats going wrong?

  1. Aug 17, 2010 #1
    Implicit differentiation, whats going wrong!?

    Hey people can someone point out to me please where I'm going wrong with part B of this question, can't get it to look the answer in the book, d2y/dx2 = -3(x^6/y^7)- 3(x^2/y^3).

    My answer is listed below under part B section, but i can't manipulate it to look like the above answer :S, any help/tips greatly appreaciated cheers,

    5. Implicit Differentiation. If x^4 + y^4 = 16, use the following steps to find y''.

    (a) Use implicit differentiation to find y',

    dy/dx = -x^(3)/y^(3) - too easy

    (b) Use the quotient or product rule to differentiate the expression for y' from part (a). Express your answer in terms of x and y only.

    d^2/dx^2= ((-x^(3))/(-3y^(4)))*y''-(3x^(2))/(y^(3))

    (c) Use the fact that x and y must satisfy x4 +y4 = 16 to simplify your answer to part

    (b) to the following expression

    d2y/dx2 = -48(x2/y7).
  2. jcsd
  3. Aug 18, 2010 #2
    Re: Implicit differentiation, whats going wrong!?

    The answer is d2y/dx2= -(3x^6/y^7) -(3x^2/y^3), cannot manipulate dy/dx = -x^3/y^3 to look like it though, any help or tips appreciated thanks,

  4. Aug 18, 2010 #3


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    Re: Implicit differentiation, whats going wrong!?


    I have no idea how you got that. In particular, I don't see how you got a y'' on the right when there was no y' on the right to begin with.

    [tex]\frac{d^2y}{dx^2}= \frac{-3x^2y^3- (-x^3)(3y^2)y'}{y^6}[/itex]
    by the quotient rule. Now, to "express your answer in terms of x and y only", replace that y' with [itex]-x^3/y^3[/itex]:
    [tex]\frac{d^2y}{dx^2}= \frac{-3x^2y^3+ 3x^3y^2(-x^3/y^3)}{y^6}[/tex]
    [tex]= \frac{-3x^2y^3- 3x^6y^{-1}}{y^6}= -3x^2\frac{y^4+ x^4}{y^7}[/itex]

    [tex]-3x^2\frac{y^4+ x^4}{y^7}[/tex]
    Last edited by a moderator: Aug 19, 2010
  5. Aug 18, 2010 #4
    Re: Implicit differentiation, whats going wrong!?

    Thanks, appreciate the help,

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