Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Review - Exercises: 58

Answer

The equation of the tangent line is $y=-1$

Work Step by Step

$y=\dfrac{x^{2}-1}{x^{2}+1}$ $,$ $(0,-1)$ Apply the quotient rule to evaluate the derivative of the given expression: $y'=\dfrac{(x^{2}+1)(x^{2}-1)'-(x^{2}-1)(x^{2}+1)'}{(x^{2}+1)^{2}}=...$ $...=\dfrac{(x^{2}+1)(2x)-(x^{2}-1)(2x)}{(x^{2}+1)^{2}}=...$ Simplify: $...=\dfrac{2x(x^{2}+1-x^{2}+1)}{(x^{2}+1)^{2}}=\dfrac{2x(2)}{(x^{2}+1)^{2}}=\dfrac{4x}{(x^{2}+1)^{2}}$ Substitute $x$ by $0$ in the derivative found to obtain the slope of the tangent line at the given point: $m=\dfrac{4(0)}{(0^{2}+1)^{2}}=\dfrac{0}{1}=0$ The slope of the tangent line and a point through which it passes are known. Use the point-slope form of the equation of a line formula, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line at the given point: $y-(-1)=(0)(x-0)$ $y+1=0$ $y=-1$ The equation of the tangent line is $y=-1$
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.