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Suppose (x^2)*(e^(-2y))= ln(xy)
Find dy/dx by implicit differentiation.

  • Math/Calculus. - ,

    2x(e^(-2y) ) + x^2(e^(-2y) )(-2dy/dx)= 1/(xy) (xdy/dx + y)

    2x e^(-2y) -2y x^2 e^(-2y) = (1/y) dy/dx + 1/x

    2x e^(-2y) - 1/x = dy/dx (1/y + 2x^2 e(-2y) )

    dy/dx = ( 2x e^(-2y) - 1/x ) / ((1/y + 2x^2 e(-2y)) )

    still pretty messy looking, I don't know how far you have to take it.

    Also, you better check my work. It is always harder to just type it here, rather than do it on paper.

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