University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 77

Answer

$$y''=2e^{x^2}(1+2x^2)$$

Work Step by Step

$$y=e^{x^2}+5x$$ - Find $y'$: $$y'=e^{x^2}(x^2)'+5=2xe^{x^2}+5$$ - Find $y''$: $$y''=2(xe^{x^2})'+(5)'=2\Big((x)'e^{x^2}+x(e^{x^2})'\Big)+0$$ $$y''=2\Big(e^{x^2}+xe^{x^2}(x^2)'\Big)=2\Big(e^{x^2}+2x^2e^{x^2}\Big)$$ $$y''=2e^{x^2}(1+2x^2)$$
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