Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 214: 24

Answer

$g'(0)=0$

Work Step by Step

Take the derivative as is on either side of the equation: $g'(x)+xcos(g(x))g'(x)+sin(g(x))=2x$ Move all terms with g'(x) onto one side of the equal sign and distribute the g'(x) out of each term: $g'(x)(1+xcos(g(x))=2x-sin(g(x))$ Isolate g'(x) by dividing both sides by the terms: $g'(x)=\frac{2x-sin(g(x))}{1+xcos(g(x)}$ Plug in 0 to find g'(0): $g'(0)=\frac{2(0)-sin(g(0))}{1+(0)cos(g(0)}=\frac{-sin(g(0))}{1}=-sin(g(0))=-sin(0)=0$ (We can see that $g(0)=0$ from the original equation.) Thus $g'(0)=0$.
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