## Calculus: Early Transcendentals 8th Edition

(a) $dy = \frac{sec^2~\sqrt{t}}{2~\sqrt{t}}~dt$ (b) $dy = \frac{-4v}{(1+v^2)^2}~dv$
(a) $y = tan~\sqrt{t}$ $\frac{dy}{dt} = sec^2~\sqrt{t}\cdot \frac{1}{2}t^{-1/2}$ $\frac{dy}{dt} = \frac{sec^2~\sqrt{t}}{2~\sqrt{t}}$ $dy = \frac{sec^2~\sqrt{t}}{2~\sqrt{t}}~dt$ (b) $y = \frac{1-v^2}{1+v^2}$ $\frac{dy}{dv} = \frac{(-2v)(1+v^2)-(2v)(1-v^2)}{(1+v^2)^2}$ $\frac{dy}{dv} = \frac{-2v-2v^3-2v+2v^3}{(1+v^2)^2}$ $\frac{dy}{dv} = \frac{-4v}{(1+v^2)^2}$ $dy = \frac{-4v}{(1+v^2)^2}~dv$