Answer
$\displaystyle \frac{dy}{dt}=ky$
(proof in step-by-step section)
Work Step by Step
$y=y_{\mathit{0}}e^{kt},\ y_{0}$ 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$:\quad \displaystyle \frac{d}{dt}(e^{g(t)})=e^{g(t)}g^{\prime}(t),\quad g(t)=kt$
$=y_{\mathit{0}}\cdot(e^{kt}\cdot k)$
$=k(y_{\mathit{0}}e^{kt})$
$=ky$
