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: 51


Equation of tangent: $x-2y+1=0$.

Work Step by Step

Given, $y=\frac{2x}{x+1}$. Differentiating $y$ with respect to $x$, at $(1,1)$, to find the slope of the tangent at the given point. Using quotient's rule, $$\frac{dy}{dx}=\frac{2(x+1) - 2x}{(x+1)^2}$$ $$\frac{dy}{dx}\bigg|_{x=1}=\frac{2(1+1) - 2\times1}{(1+1)^2}=\frac{1}{2}.$$ Using point-slope form of a line, the equation of tangent through $(1,1)$ is, $$y-1=\frac{1}{2}(x-1)$$ $$2y-2=x-1\implies x-2y+1=0.$$
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.