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Thread: Fidn x^2y+xy^2=6

  1. #1
    Mar 2011

    Fidn x^2y+xy^2=6

    x^2y+xy^2=6. Find d^2y/dx^2 at (1,3). And if it equals A) -18 B) -6 C)6 D) 12 E)18?

    2) x^3+y^3=27. find d^2y/dx^2
    Please help me with this calculus problem. I am so lost in my class. step by step work would be greatly appreciated. The work does not have to be completely accurate, I just need the steps. Thank you in advance.
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  2. #2
    MHF Contributor

    Apr 2005
    Use "implicit differentiation" which is basically using the chain rule. Differentiate both sides of the equation $\displaystyle x^2y+ xy^2= 6$ with respect to x: [tex](x^2y+ xy^2)'= (x^2y)'+ (xy^2)'. Use the product rule on $\displaystyle (x^2y)'= (x^2)'y+ x^2y'= 2xy+ x^2y'$. Similarly, $\displaystyle (xy^2)'= (x')y^2+ x(y^2)'= y^2+ x(2y y')$ (that last term is the chain rule). Of course, the derivative of any constant is 0 so the derivative on the right is (6)'= 0. Putting those together,
    $\displaystyle 2xy+ x^2y'+ y^2+ 2xy y'= 0$. You can x= 1, y= 3 to get an equation to solve for y' at that point.

    Now go ahead and use the product and chain rules to differentiate again.
    $\displaystyle 2y+ 2xy'+ 2xy'+ x^2y''+ 2yy'+ 2y y'+ 2x y'^2+ 2xy y''= 0$. Put x= 1, y= 3, and y' equal to whatever you got above and solve the equation for y''.
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