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: 40


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

Work Step by Step

Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(x^n)'=nx^{n-1}$ Since $ y=\frac{1}{(1-x)\sqrt{2-x}}$, we rewrite $ y $ as follows $$ y=\frac{1}{(1-x)\sqrt{2-x}}= (1-x)^{-1}(2-x)^{-1/2}.$$ Now, the derivative $ y'$, by using the product rule, is given by $$ y'=(-1)(-1)(1-x)^{-2}(2-x)^{-1/2}-\frac{1}{2}(-1)(1-x)^{-1}(2-x)^{-3/2} = (1-x)^{-2}(2-x)^{-1/2}+\frac{1}{2}(1-x)^{-1}(2-x)^{-3/2} .$$
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