# Calculus (repost)

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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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