Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 62


$$ y'= \frac{(1+x^2)\sin(1/x)-2x^3\cos(1/x)}{x^2(1+x^2)^2}.$$

Work Step by Step

Since $ y=\frac{\cos(1/x)}{1+x^2}$, by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ and by using the quotient rule; $(u/v)'=\frac{vu'-uv'}{v^2}$ and recalling that $(\cos x)'=-\sin x $, the derivative $ y'$ is given by $$ y'=\frac{-(1+x^2)\sin(1/x) (1/x)'-2x\cos(1/x)}{(1+x^2)^2} \\=\frac{-(-1/x^2)(1+x^2)\sin(1/x)-2x\cos(1/x)}{(1+x^2)^2}\\ =\frac{(1+x^2)\sin(1/x)-2x^3\cos(1/x)}{x^2(1+x^2)^2}.$$
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