University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 62


$$y'=e^{\sin t}\Big(\cos t(\ln t^2+1)+\frac{2}{t}\Big)$$

Work Step by Step

$$y=e^{\sin t}(\ln t^2+1)$$ We have $$(\ln t)'=\frac{1}{t}$$ $$(e^{t})'=e^t$$ We also need to apply the Product Rule and the Chain Rule here. The derivative of $y$ is $$y'=e^{\sin t}(\sin t)'(\ln t^2+1)+e^{\sin t}\Big(\frac{1}{t^2}(t^2)'+0\Big)$$ $$y'=e^{\sin t}\cos t(\ln t^2+1)+e^{\sin t}\frac{2t}{t^2}$$ $$y'=e^{\sin t}\cos t(\ln t^2+1)+\frac{2e^{\sin t}}{t}$$ $$y'=e^{\sin t}\Big(\cos t(\ln t^2+1)+\frac{2}{t}\Big)$$
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