Calculus (3rd Edition)

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

Chapter 3 - Differentiation - 3.1 Definition of the Derivative - Exercises - Page 103: 27


$$\frac{ -1}{10\sqrt{5}}$$

Work Step by Step

Since $ f(x)=\frac{1}{\sqrt{x}}$, we have $$\frac{f(5+h)-f(5)}{h}=\frac{\frac{1}{\sqrt{5+h}}-\frac{1}{\sqrt{5}}}{h}=\frac{ \sqrt{5}-\sqrt{5+h}}{h\sqrt{5}\sqrt{5+h}}.$$ Now, rationlizing the above fraction by multiplying by $\frac{\sqrt{5}+\sqrt{5+h}}{\sqrt{5}+\sqrt{5+h}}$, we get $$ \frac{ \sqrt{5}-\sqrt{5+h}}{h\sqrt{5}\sqrt{5+h}} \frac{\sqrt{5}+\sqrt{5+h}}{\sqrt{5}+\sqrt{5+h}}=\frac{ 5-(5+h)}{h\sqrt{5}\sqrt{5+h}(\sqrt{5}+\sqrt{5+h})} .$$ Finally, we get $$\frac{f(5+h)-f(5)}{h}=-\frac{ 1}{\sqrt{5}\sqrt{5+h}(\sqrt{5}+\sqrt{5+h})} .$$ Plugging in $h=0$, we get: $$-\frac{ 1}{\sqrt{5}\sqrt{5}(\sqrt{5}+\sqrt{5})} =\frac{ -1}{5(2\sqrt{5})}$$ $$=\frac{ -1}{10\sqrt{5}}$$
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