Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises - Page 141: 53


(a) Equation of tangent: $x-2y+2=0$. (b) Check the graph given below

Work Step by Step

(a) $y=\frac{1}{1+x^2}$. Differentiating $y$ with respect to $x$ at $(-1,1/2)$. Using product's rule, $$\frac{dy}{dx}=\frac{d}{dx}(1+x^2)^{-1}=(-1)(1+x^2)^{-2}(2x)$$ $$\frac{dy}{dx}\bigg|_{x=-1}=(-1)(1+(-1)^2)^{-2}(2\times-1)=-1\times\frac{1}{2^2}\times-2=\frac{1}{2}.$$ Thus, the slope of the tangent line is $1/2$. Using point-slope form, $$y-\frac{1}{2}=\frac{1}{2}(x+1)$$ $$\frac{2y-1}{2}=\frac{1}{2}(x+1)\implies 2y-1=x+1$$ $$x-2y+2=0$$ is the equation of the required tangent. (b) Check the graph given below
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