Answer
$\frac{(e^x)(x^2-1)-(e^x)(2x)}{(x^2-1)^2}$
Work Step by Step
$\left(\frac{f}{g}\right)'$ = $\frac{f'g-fg'}{g^2}$
$\left(\frac{e^x}{x^2-1}\right)'$ = $\frac{(e^x)(x^2-1)-(e^x)(2x)}{(x^2-1)^2}$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises - Page 160: 27
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