Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises: 31

Answer

$$z'=\frac{-10A}{y^{11}}+Be^y$$

Work Step by Step

$$z=\frac{A}{y^{10}}+Be^y$$ $$z=Ay^{-10}+Be^y$$ So, the derivative of $z$ is $$z'=A\frac{d}{dy}(y^{-10})+B\frac{d}{dy}(e^y)$$ $$z'=A\times(-10)y^{-11}+Be^y$$ $$z'=-10Ay^{-11}+Be^y$$ $$z'=\frac{-10A}{y^{11}}+Be^y$$
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