Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 327: 44


$\frac{d^2y}{dt^2}=e^{-2t}(-5\sin3t-12\cos 3t).$

Work Step by Step

Since we have $$ y= e^{-2t} \sin 3t $$ then the first derivative $\frac{dy}{dt}$, using the product rule, is given by \begin{align*} \frac{dy}{dt}&= (e^{-2t})' \sin3t+e^{-2t} (\sin 3t)'=-2e^{-2t} \sin3t+3e^{-2t} \cos 3t\\&=e^{-2t}( -2\sin 3t+ 3\cos 3t).\end{align*} The second derivative $\frac{d^2y}{dt^2}$ is given by \begin{align*}\frac{d^2y}{dt^2}&=(e^{-2t})'( -2\sin 3t+3\cos 3t)+e^{-2t}( -2\sin 3t+3\cos 3t)' \\ &=-2e^{-2t}( -2\sin 3t+3\cos 3t)+e^{-2t}(-6\cos 3t-9\sin3t)\\ &=e^{-2t}(-5\sin3t-12\cos 3t). \end{align*}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.