Answer
$y'=\dfrac{2x-y\cos(xy)}{x\cos(xy)+1}$
Work Step by Step
$\sin(xy)=x^{2}-y$
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:
$y'=\dfrac{2x-y\cos(xy)}{x\cos(xy)+1}$
