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: 39



Work Step by Step

$$\frac{dy}{dx}=\Big(\frac{(2x+3)^3}{(4x^2-1)^8}\Big)'= \frac{((2x+3)^3)'(4x^2-1)^8-(2x+3)^3((4x^2-1)^8)'}{(4x^2-1)^{16}}= \frac{3(2x+3)^2(2x+3)'(4x^2-1)^8-(2x+3)^3\cdot8(4x^2-1)^7(4x^2-1)'}{(4x^2-1)^{16}}= \frac{6(2x+3)^2(4x^2-1)^8-64x(2x+3)^3(4x^2-1)^7}{(4x^2-1)^{16}}= \frac{6(2x+3)^2(4x^2-1)-64x(2x+3)^3}{(4x^2-1)^9}= \frac{2(2x+3)^2(12x^2-3-64x^2-96x)}{(4x^2-1)^9} =-\frac{2(2x+3)^2(52x^2+96x+3)}{(4x^2-1)^9}$$
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