## University Calculus: Early Transcendentals (3rd Edition)

$\frac{dy}{dt} = -6$
Given that $\frac{dx}{dt} = 3$ and $y=x^2$ on differentiating y with respect to time $\frac{d(y)}{dt}$ = $\frac{d x^2}{dt}$ $\frac{dy}{dt} = \frac{2xdx}{dt}$ so $\frac{dy}{dt} = 2x\times 3$ $\frac{dy}{dt} = 6x$ when x=-1 then $\frac{dy}{dt} = 6\times (-1)$ Thus, the answer is $\frac{dy}{dt} = -6$