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

Answer

(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
Small 1559322870
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.