Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises - Page 142: 75

Answer

\[(a) y'=g(x)+xg'(x)\] \[(b) y'=\frac{g(x)-xg'(x)}{[g(x)]^2}\] \[(c) y'=\frac{xg'(x)-g(x)}{x^2}\]

Work Step by Step

$(a)\; y=xg(x)$ Differentiating with respect to $x$ by product rule $y'=(x)'g(x)+xg'(x)$ $y'=g(x)+xg'(x)$ Hence $\;y'=g(x)+xg'(x)$. $(b) \; y=\frac{x}{g(x)}$ Differentiating with respect to $x$ by quotient rule $y'=\frac{(x)'g(x)-xg'(x)}{[g(x)]^2}$ $y'=\frac{g(x)-xg'(x)}{[g(x)]^2}$ Hence $\;y'=\frac{g(x)-xg'(x)}{[g(x)]^2}$. $(c) \; y=\frac{g(x)}{x}$ Differentiating with respect to $x$ by quotient rule $y'=\frac{xg'(x)-g(x)(x)'}{x^2}$ $y'=\frac{xg'(x)-g(x)}{x^2}$ Hence $\;y'=\frac{xg'(x)-g(x)}{x^2}$.
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