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use implicit differentiation to find y' if x^3-xy+y^3=7
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3x^2 - y - xy' + 3y^2y' = 0 Now solve for y'. Your answer will have x's and y's in it.
Mark do you understand how bang came up with those middle two terms? It requires the product rule:\[\Large (-xy)'\quad=\quad -(x)'y-x(y)'\quad=\quad -(1)y-x(y')\]
Yeah, i think the product rule with implicit differentiation confuses a lot of people.
how do i get rid of the y' in the exponent
Hmm there are no y' in exponent positions.\[\Large (y^3)'\quad=\quad (3y^2)y'\]
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I got it now i was reading it wrong before
Oh ok :)
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