Thomas' Calculus 13th Edition

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

Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 69

Answer

$0$

Work Step by Step

$(fog)'(x)$ = $f'(g(x))g'(x)$ $g(x)$ = $10x^{2}+x+1$ $g(0)$ = $10(0)^{2}+0+1$ = $1$ $g'(x)$ = $20x+1$ $g'(0)$ = $20(0)+1$ = $1$ $f(u)$ = $\frac{2u}{u^{y2}+1}$ $f'(u)$ = $\frac{d}{dx}$[$\frac{2u}{u^{2}+1}$] $f'(u)$ = $\frac{({u^{2}+1})\frac{d}{dx}(2u)-(2u)\frac{d}{dx}({u^{2}+1})}{({u^{2}+1})^{2}}$ $f'(u)$ = $\frac{({u^{2}+1})(2)-(2u)({2u})}{({u^{2}+1})^{2}}$ = $\frac{2u^{2}+2-4u^{2}}{({u^{2}+1})^{2}}$ = $\frac{-2u^{2}}{({u^{2}+1})^{2}}$ $f'(1)$ = $f'(g(0))$ = $\frac{-2u^{2}+2}{({u^{2}+1})^{2}}$ = $\frac{-2(1)^{2}+2}{({(1)^{2}+1})^{2}}$ = $0$ $(fog)'(0)$ = $f'(g(0))g'(0)$ = $(0)$$(1)$ = $0$
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