Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.4 Derivatives of Exponential Functions - 4.4 Exercises - Page 232: 35


$\displaystyle \frac{dy}{dt}=ky$ (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$:\quad \displaystyle \frac{d}{dt}(e^{g(t)})=e^{g(t)}g^{\prime}(t),\quad g(t)=kt$ $=y_{\mathit{0}}\cdot(e^{kt}\cdot k)$ $=k(y_{\mathit{0}}e^{kt})$ $=ky$
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