Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.3 Differentiation Rules - Exercises 3.3: 20

Answer

The Derivative is: $y'(t)=(t\space+\space2)^{-2}$

Work Step by Step

$f(t)=\frac{t^2-1}{t^2+t-2}$ Simplifying the function $f(t)=\frac{(t+1)\cdot (t-1)}{(t-1)\cdot(t+2)}$ $f(t)=\frac{t+1}{t+2}$ Using the Quotient Rule to find the Derivative: $y'(t)=\frac{f'(t)\cdot g(t)\space - \space f(t)\cdot g'(t)}{g^2(t)}$ $y'(t)=\frac{((1)t^{1-1}+\space 0)(t\space +\space 2)\space -\space (t\space+\space 1)((1)t^{1-1}+\space 0)}{(t\space+\space 2)^2}$ $y'(t)=\frac{(1)(t+2)\space -\space (t+1)(1)}{(t\space+\space 2)^2}$ $y'(t)=\frac{(t+2)\space -\space (t+1)}{(t\space+\space 2)^2}$ $y'(t)=\frac{t\space+\space2\space-\space t\space-\space1}{(t\space+\space2)^2}$ $y'(t)=\frac{1}{(t\space+\space2)^2}$ OR $y'(t)=(t\space+\space2)^{-2}$
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.