# Implicit differentiation for $y\cos x = 4x^2 + 3y^2$

I am stuck on doing this implicit differentiation problem below. $$y \cos x = 4x^2 + 3y^2$$

I am now stuck at the following equality and I don't know how to proceed. Can someone help me? $$y(−\sin x) + (\cos x)y' = 8x + 6yy'$$

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What does the problem ask you to do? If you're trying to solve for $y'$ in terms of $x$ and $y$, try getting both $y'$ terms on one side of the equation and factoring. – Jonas Meyer Jun 29 '12 at 20:36
Yeah im trying to find $y'$ – soniccool Jun 29 '12 at 20:37
@soniccool Is it clear how you get $$y(− \sin x) + \cos x · y' = 8x + 6yy'$$ from $$y \cos x = 4x^2 + 3y^2?$$ – user17762 Jun 29 '12 at 20:38

Group the $y'$ terms to one side to get $$(\cos(x)-6y)y' = 8x + y \sin(x) \implies y' = \dfrac{8x + y \sin(x)}{\cos(x) - 6y}$$
What Marvis noted can be formulated below. Let $F(x,y)=0$ is a relation linking two variables $x$, $y$ implicitly. Then you can find $y'$ from this relation by calculating $$y'=\frac{-F_x}{F_y}$$ where in $F_x$ is differentiate of $F$ with respect to $x$. Now, $y \cos x = 4x^2 + 3y^2$ becames $y \cos x - 4x^2 - 3y^2=0$. Take $F(x,y)=y \cos x - 4x^2 - 3y^2=0$ and the rest is as Marvis did completely.