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: 45

Answer

\begin{align*}\frac{d^2y}{dt^2}&= \left(4 t^{2}-4 t-1\right) e^{t-t^{2}}. \end{align*}

Work Step by Step

Since we have $$ y= e^{t-t^2} $$ then the first derivative $\frac{dy}{dt}$, using the chain rule, is given by \begin{align*} \frac{dy}{dt}&= e^{t-t^2} (t-t^2)'\\ &=(1-2t)e^{t-t^2} \end{align*} The second derivative $\frac{d^2y}{dt^2}$, using the product rule, is given by \begin{align*}\frac{d^2y}{dt^2}&=(1-2t)'e^{t-t^2}+(1-2t)(e^{t-t^2})' \\ &=(-2)e^{t-t^2}+(1-2t)(1-2t)e^{t-t^2}\\ &= \left(4 t^{2}-4 t-1\right) e^{t-t^{2}}. \end{align*}
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