Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6: 37

Answer

$$\frac{dy}{dx}=\frac{33(x-5)^2}{(2x+1)^4}$$

Work Step by Step

$$\frac{dy}{dx}=\Big(\Big(\frac{x-5}{2x+1}\Big)^3\Big)'=3\Big(\frac{x-5}{2x+1}\Big)^2\Big(\frac{x-5}{2x+1}\Big)'= 3\Big(\frac{x-5}{2x+1}\Big)^2\frac{(x-5)'(2x+1)-(x-5)(2x+1)'}{(2x+1)^2}= 3\Big(\frac{x-5}{2x+1}\Big)^2\frac{2x+1-2x+10}{(2x+1)^2}= 3\Big(\frac{x-5}{2x+1}\Big)^2\frac{11}{(2x+1)^2}=\frac{33(x-5)^2}{(2x+1)^4}$$
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