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Hello all

If we are given [tex] \cos xy = 2x^2 - 3y [/tex] find [tex] \frac {dy}{dx} [/tex]

So the derivative of [tex] \cos xy [/tex] is [tex] - sin(xy)(x)(\frac{dy}{dx} + y) [/tex]

The derivative of the RHS is [tex] 4x - 3 \frac {dy}{dx} [/tex]

Hence [tex] \-sin(xy)(x)\frac{dy}{dx} + y [/tex] = [tex] 4x - 3 \frac {dy}{dx} [/tex]

HOw do I find [tex] \frac {dy}{dx} [/tex] ?

Is the answer [tex] \frac {dy}{dx} = \frac{y\sin xy + 4x}{ -x\sin xy + 3} [/tex]

Thanks

(sin should be (- sin)

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# Homework Help: Verification of Implicit Differentiation

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