Calculus AP
posted by karen .
implicit differentiation
Find the slope of the tangent line to the curve xy^3 + 2y  0.76 =0 at the point (3, o.7)
I'm lost.
so 3xy^2 * y' + 2y = 0 right?

not quite.
xy^3 + 2y  0.76 =0
y^3 + 3xy^2 y' + 2y' = 0
y' = y^3/(3xy^2 + 2)
xy^3 is a product f*g
(f*g(' = f'g + fg'
at (3,0.7) y' = .7^3/(3(3)(.7^2) + 2)
= . . .
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