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Use implicit differentiation to find the slope of the tangent line to the curve 3 x^2 - 4 xy + 3 y^3 = 4 at the point ( 2,-2 ) how can you solve this?
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Solve for y'(x) \[ 9 y(x)^2 y'(x)-4 x y'(x)-4 y(x)+6 x=0 \]
\[ y'(x)=\frac{2 (3 x-2 y(x))}{4 x-9 y(x)^2} \]
thank you :D
YW
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