## Calculus with Applications (10th Edition)

$\displaystyle \frac{dy}{dt}=ky$ (proof in step-by-step section)
$y=y_{\mathit{0}}e^{kt},\ y_{0}$ and k constants, $\displaystyle \frac{dy}{dt}=\frac{d}{dt}[y_{\mathit{0}}e^{kt}]$ ... $y_{0}$ is a constant (multiplying a function), $=y_{0}\displaystyle \cdot\frac{d}{dt}[e^{kt}]$ ... use$:\quad \displaystyle \frac{d}{dt}(e^{g(t)})=e^{g(t)}g^{\prime}(t),\quad g(t)=kt$ $=y_{\mathit{0}}\cdot(e^{kt}\cdot k)$ $=k(y_{\mathit{0}}e^{kt})$ $=ky$