## Calculus (3rd Edition)

$$\frac{d}{dx}\tan(xy)= (y+x\frac{dy}{dx})\sec^2 (xy).$$
Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(\tan x)'=\sec^2 x$. Using the chain rule, we have $$\frac{d}{dx}\tan(xy)= (y+x\frac{dy}{dx})\sec^2 (xy).$$