Implicit differentiation of sec(x+y) = xy

Differentiate both sides with respect to $x$. The left side requires the chain rule: $\dfrac{d}{dx}[\sec(x+y)] = \sec(x+y)\tan(x+y)\cdot\left(1 + \dfrac{dy}{dx}\right)$, since the derivative of the inner function $x+y$ is $1 + \dfrac{dy}{dx}$. The right side requires the product rule: $\dfrac{d}{dx}[xy] = y + x\dfrac{dy}{dx}$.

Setting the two sides equal gives $\sec(x+y)\tan(x+y)\left(1+\frac{dy}{dx}\right) = y + x\frac{dy}{dx}.$ Expand the left side, then move all terms with $\dfrac{dy}{dx}$ to one side: $\sec(x+y)\tan(x+y) - y = x\frac{dy}{dx} - \sec(x+y)\tan(x+y)\frac{dy}{dx}.$ Factor out $\dfrac{dy}{dx}$ and divide to solve: $\frac{dy}{dx} = \frac{\sec(x+y)\tan(x+y) - y}{x - \sec(x+y)\tan(x+y)}.$