Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises - Page 205: 57

Answer

(a) $f'(x) = \frac{2-2x^2}{\sqrt{2-x^2}}$ (b) We can see a sketch of the graphs below.
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Work Step by Step

(a) $f(x) = x~\sqrt{2-x^2}$ We can find $f'(x)$: $f'(x) = \sqrt{2-x^2}+\frac{1}{2}(2-x^2)^{-1/2}(-2x)(x)$ $f'(x) = \frac{2-x^2}{\sqrt{2-x^2}}-\frac{x^2}{\sqrt{2-x^2}}$ $f'(x) = \frac{2-2x^2}{\sqrt{2-x^2}}$ (b) We can see a sketch of the graphs below. $f'(x)$ seems reasonable since $f'(x)$ has negative values when the slope of $f(x)$ has a negative slope, $f'(x)$ is 0 when the slope of $f(x)$ is 0, and $f'(x)$ has positive values when the slope of $f(x)$ has a positive slope.
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