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: 34



Work Step by Step

$$y=(2x-5)^{-1}(x^2-5x)^6$$ The derivative of $y$ is: $$y'=\Big((2x-5)^{-1}\Big)'(x^2-5x)^6+(2x-5)^{-1}\Big((x^2-5x)^6\Big)'$$ $$y'=-(2x-5)^{-2}(2x-5)'(x^2-5x)^6+(2x-5)^{-1}\Big(6(x^2-5x)^5(x^2-5x)'\Big)$$ $$y'=-2(2x-5)^{-2}(x^2-5x)^6+6(2x-5)^{-1}(x^2-5x)^5(2x-5)$$ $$y'=-2(2x-5)^{-2}(x^2-5x)^6+6(x^2-5x)^5$$
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