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Calculus (repost)

posted by on .

Use implicit differentiation to find dy/dx if cos xy = 2x^2 - 3y.

I'm stuck on this problem because I'm getting thrown off on how to factor this. Here's my work so far:

-{[(dy/dx)y + y]sin xy} = 4x - 3(dy/dx)
-{[(dy/dx)y + y]sinxy} + 3(dy/dx) = 4x

...now what?

Thx

  • Calculus (repost) - ,

    factor out dy/dx which is what you want

    dy/dx [ 3 -y sinxy] = [ 4 x + y sin xy ]
    so
    dy/dx =[ 4 x + y sin xy ] / [ 3 -y sinxy]

  • Calculus (repost) - ,

    There is a minor correction to the original differentiation:
    -{[(dy/dx)x + y]sin xy} = 4x - 3(dy/dx)

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