Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.4 - Rational Expressions - 1.4 Exercises - Page 43: 74

Answer

$\dfrac{\dfrac{1}{\sqrt{x+h}}-\dfrac{1}{\sqrt{x}}}{h}=\dfrac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x^{2}+xh}}$

Work Step by Step

$\dfrac{\dfrac{1}{\sqrt{x+h}}-\dfrac{1}{\sqrt{x}}}{h}$ Evaluate the subtraction in the numerator: $\dfrac{\dfrac{1}{\sqrt{x+h}}-\dfrac{1}{\sqrt{x}}}{h}=\dfrac{\dfrac{\sqrt{x}-\sqrt{x+h}}{(\sqrt{x})(\sqrt{x+h})}}{h}=...$ Evaluate the division: $...=\dfrac{\sqrt{x}-\sqrt{x+h}}{h(\sqrt{x})(\sqrt{x+h})}=...$ Evaluate the product in the denominator: $...=\dfrac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x^{2}+xh}}$
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