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: 21

Answer

The Derivative is: $v'=\frac{t^2-2t-1}{(1+t^2)^2}$

Work Step by Step

$v=\left(1-t\right)\space (1+t^2)^{-1}$ $v=\frac{1\space-\space t}{1\space+\space t^2}$ Using the Quotient Rule to find the Derivative: $v'=\frac{f'(t)\cdot g(t)\space-\space f(t)\cdot g'(t)}{g^2(t)}$ $v'=\frac{(0\space-\space(1)t^{1-1})(1\space+\space t^2)\space-\space(1\space-\space t)(0\space+\space(2)t^{2-1})}{(1\space+\space t^2)^2}$ $v'=\frac{(-1)(1+t^2)-(1-t)(2t)}{(1+t^2)^2}$ $v'=\frac{-1-t^2-2t+2t^2}{(1+t^2)^2}$ $v'=\frac{t^2-2t-1}{(1+t^2)^2}$
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