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Q. 48

Expert-verified

Calculus

Found in: Page 184

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Use the definition of the derivative to find $f'$ for each function $f$ in Exercises
$f (x) = \frac{3}{\sqrt{x}}$

The value of $f' (x) = \frac{- 9}{6 x^{\frac{3}{2}}}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information

The given function $f (x) = \frac{3}{\sqrt{x}}$

Step 2. Finding the value of f'(x)

We know that $f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} . . . . . (1)$

Given $f (x) = \frac{3}{\sqrt{x}}$

Then $f (x + h) = \frac{3}{\sqrt{x + h}}$

Putting these values in (1)

$f' (x) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{x + h}} - \frac{3}{\sqrt{x}}}{h} = \lim_{h \to 0} \frac{\frac{3 \sqrt{x} - 3 \sqrt{x + h}}{\sqrt{x} \sqrt{x + h}}}{h} = \lim_{h \to 0} \frac{\frac{3 \sqrt{x} - 3 \sqrt{x + h}}{\sqrt{x} \sqrt{x + h}} \times \frac{3 \sqrt{x} + 3 \sqrt{x + h}}{3 \sqrt{x} + 3 \sqrt{x + h}}}{h} = \lim_{h \to 0} \frac{\frac{{(3 \sqrt{x})}^{2} - {(3 \sqrt{x + h})}^{2}}{\sqrt{x} \sqrt{x + h} (3 \sqrt{x} + 3 \sqrt{x + h})}}{h} = \lim_{h \to 0} \frac{\frac{9 x - 9 x - 9 h}{\sqrt{x} \sqrt{x + h} (3 \sqrt{x} + 3 \sqrt{x + h})}}{h} = \lim_{h \to 0} \frac{\frac{- 9 h}{\sqrt{x} \sqrt{x + h} (3 \sqrt{x} + 3 \sqrt{x + h}}}{h} = \lim_{h \to 0} \frac{- 9}{\sqrt{x} \sqrt{x + h} (3 \sqrt{x} + 3 \sqrt{x + h})}$

Putting $h = 0$

$= \frac{- 9}{\sqrt{x} \sqrt{x + 0} (3 \sqrt{x} + 3 \sqrt{x + 0})} = \frac{- 9}{\sqrt{x} \sqrt{x} (3 \sqrt{x} + 3 \sqrt{x})} = \frac{- 9}{x (2.3 \sqrt{x})} = \frac{- 9}{6 x^{\frac{3}{2}}}$

Hence,the value of $f' (x) = \frac{- 9}{6 x^{\frac{3}{2}}}$

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