Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.3 Differentiation Rules - Exercises 3.3 - Page 125: 22


The Derivative is: $w'=-\frac{17}{(2x-7)^2}$

Work Step by Step

$w=(2x-7)^{-1}(x+5)$ $w=\frac{x+5}{2x-7}$ Using Quotient Rule to find the Derivative: $w'=\frac{f'(x)\cdot g(x) - f(x)\cdot g'(x)}{g^2(x)}$ $w'=\frac{((1)x^{1-1}+0)(2x-7)-(x+5)(2x^{1-1}-0)}{(2x-7)^2}$ $w'=\frac{(1)(2x-7)-(x+5)(2)}{(2x-7)^2}$ $w'=\frac{2x-7-2x-10}{(2x-7)^2}$ $w'=-\frac{17}{(2x-7)^2}$
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