Jonathan W. answered • 02/18/14

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I can get you started with the partial derivative ∂R/∂x = abx^(a-1)y^b. Have you gotten that far?

Garrick N.

Nevermind, I've got it now.

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02/18/14

Garrick N.

Am I to just add ∂R/∂x = abx^(a-1)y^b and ∂R/∂y = b^2x^ay^(b-1) together for the first part?

And, I'm not sure where to start with the second part.

02/18/14