Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 55

Answer

$$R'(0)=1$$

Work Step by Step

$$R(x)=\frac{x-3x^3+5x^5}{1+3x^3+6x^6+9x^9}$$ $$R(x)=\frac{f(x)}{g(x)}$$ According to Quotient Rule, $$R'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$$ Therefore, $$R'(0)=\frac{f'(0)g(0)-f(0)g'(0)}{[g(0)]^2}$$ *Find $f(0)$, $g(0)$, $f'(0)$ and $g'(0)$ $$f(x)=x-3x^3+5x^5$$ So, $$f'(x)=1-9x^2+25x^4$$ Therefore, $f(0)=0-3\times0^3+5\times0^5=0$ and $f'(0)=1-9\times0^2+25\times0^4=1$ $$g(x)=1+3x^3+6x^6+9x^9$$ So, $$g'(x)=9x^2+36x^5+81x^8$$ Therefore, $g(0)=1+3\times0^3+6\times0^6+9\times0^9=1$ and $g'(0)=9\times0^2+36\times0^5+81\times0^8=0$ That means $$R'(0)=\frac{1\times1-0\times0}{1^2}=1$$
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