## Calculus, 10th Edition (Anton)

$y' = \dfrac{1-2xy^2 \cos(x^2y^2)}{2x^2y\cos (x^2y^2)}$
In order to derivate this function you have to apply implicit differentation method. First, take the function to it's f(x)=0 form $\sin(x^2y^2)-x=0$ Then derivate the whole equation. Rember the put y' every time you derivate y $(2xy^2+2x^2yy')\cos(x^2y^2)-1=0$ *Note: In this one, you have to apply the chain rule Solve for y' and you have the answer $y'(2x^2y\cos (x^2y^2))=1-2xy^2 \cos(x^2y^2)$ $y' = \dfrac{1-2xy^2 \cos(x^2y^2)}{2x^2y\cos (x^2y^2)}$