## Calculus (3rd Edition)

Implicitly differentiating $xy=1$ with respect to $t$, we have $\frac{dx}{dt}\times y+ x\frac{dy}{dt}=0$ $\implies x \frac{dy}{dt}=-y\frac{dx}{dt}$ $\implies \frac{dy}{dt}=-\frac{y}{x}\frac{dx}{dt}$