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: 37



Work Step by Step

$$y=(x^2-2x+2)e^{5x/2}$$ The derivative of function $y$ is: $$y'=(x^2-2x+2)'e^{5x/2}+(x^2-2x+2)(e^{5x/2})'$$ $$y'=(2x-2)e^{5x/2}+(x^2-2x+2)e^{5x/2}\Big(\frac{5x}{2}\Big)'$$ $$y'=(2x-2)e^{5x/2}+\frac{5}{2}(x^2-2x+2)e^{5x/2}$$ $$y'=e^{5x/2}\Big(2x-2+\frac{5}{2}x^2-5x+5\Big)$$ $$y'=e^{5x/2}\Big(\frac{5}{2}x^2-3x+3\Big)$$
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