Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - Chapter Review Exercises - Page 163: 37


$$ y'=\frac{ 1 -\frac{z}{2}}{(1-z)^{3/2}}.$$

Work Step by Step

Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(x^n)'=nx^{n-1}$ Since $ y= \frac{z}{\sqrt{1-z}}$, then by the quotient rule the derivative $ y'$ is given by $$ y'=\frac{\sqrt{1-z} +\frac{z}{2\sqrt{1-z}}}{1-z} =\frac{ 1-z +\frac{z}{2}}{(1-z)\sqrt{1-z}}=\frac{ 1 -\frac{z}{2}}{(1-z)^{3/2}}.$$
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