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 - Page 158: 40



Work Step by Step

$$\frac{dy}{dx}=((1+\sin^3(x^5))^{12})'= 12(1+\sin^3(x^5))^{11}(1+\sin^3(x^5))'= 12(1+\sin^3(x^5))^{11}(3\sin^2(x^5)\cdot(\sin(x^5))')= 36(1+\sin^3(x^5))^{11}\sin^2(x^5)\cos(x^5)(x^5)'= 180x^4(1+\sin^3(x^5))^{11}\sin^2(x^5)\cos(x^5)= 90x^4(1+\sin^3(x^5))^{11}\sin(x^5)\sin(2x^5)$$
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