## Calculus: Early Transcendentals 8th Edition

$y'=\dfrac{2x-y\cos(xy)}{x\cos(xy)+1}$
$\sin(xy)=x^{2}-y$ Implicit differentiation must be used for this exercise. Apply the chain rule to both sides of the equation to evaluate the derivative: $\cos(xy)(xy)'=2x-y'$ Apply the product rule to evaluate the indicated derivative: $\cos(xy)(xy'+y)=2x-y'$ Evaluate the product indicated on the left side of the equation: $xy'\cos(xy)+y\cos(xy)=2x-y'$ Take $y'$ to the left side and $y\cos(xy)$ to the right side: $xy'\cos(xy)+y'=2x-y\cos(xy)$ Take out common factor $y'$ from the left side: $y'[x\cos(xy)+1]=2x-y\cos(xy)$ Solve for $y'$ to finish the differentiation process: $y'=\dfrac{2x-y\cos(xy)}{x\cos(xy)+1}$