Calculus 8th Edition

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

Chapter 2 - Derivatives - 2.1 Derivatives and Rates of Change - 2.1 Exercises: 8

Answer

$y=\frac{1}{3}x+\frac{2}{3}$

Work Step by Step

$y=\frac{2x+1}{x+2} ,(1,1)$ 1. Find the first derivative by using the quotient rule. Finding the first derivative gives you the slope of the tangent line. $y'=\frac{(x+2)(2)-(2x+1)(1)}{(x+2)^2}$ 2. Plug in the given x value to find the slope of the tangent line to the curve at the given point. There is no need to simplify since we are looking for the equation of the tangent line, not the equation for the slope. $y'=\frac{(1+2)(2)-(2+1)(1)}{(1+2)^2}$ $y'=\frac{1}{3}$ 3. Now that we have the slope of the tangent line, we can write the equation for that line in point-slope form. $y-y_{1}=m(x-x_{1})$ $y-1=\frac{1}{3}(x-1)$ 4. Rewrite into standard form by distributing the 1/3 and adding one to both sides. $y-1+1=\frac{1}{3}x-\frac{1}{3}+1$ 5. Clean up and simplify. $y=\frac{1}{3}x+\frac{2}{3}$
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