Chapter 2: Q. 45 (page 184)

Use the definition of the derivative to find $f'$ for each function $f$ in Exercises 34-59
$f (x) = \frac{1}{x^{2}}$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information

The given function $f (x) = \frac{1}{x^{2}}$

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 $f (x) = \frac{1}{x^{2}}$ the

Putting the values in (1)

localid="1648748944158" role="math" $f' (x) = \lim_{h \to 0} \frac{\frac{1}{{(x + h)}^{2}} - \frac{1}{x^{2}}}{h} = \lim_{h \to 0} \frac{\frac{x^{2} - {(x + h)}^{2}}{x^{2} {(x + h)}^{2}}}{h} = \lim_{h \to 0} \frac{\frac{x^{2} - x^{2} - h^{2} - 2 x h}{x^{2} {(x + h)}^{2}}}{h} = \lim_{h \to 0} \frac{\frac{- h^{2} - 2 x h}{x^{2} {(x + h)}^{2}}}{h} = \lim_{h \to 0} \frac{\frac{h (- h - 2 x)}{x^{2} {(x + h)}^{2}}}{h} = \lim_{h \to 0} \frac{h (- h - 2 x)}{x^{2} {(x + h)}^{2}} \times \frac{1}{h} = \lim_{h \to 0} \frac{(- h - 2 x)}{x^{2} {(x + h)}^{2}}$

Putting $h = 0$

$= \frac{- (0) - 2 x}{x^{2} . {(x + 0)}^{2}} = \frac{- 2 x}{x^{2} . x^{2}} = \frac{- 2}{x^{3}}$

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

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Use the definition of the derivative to find $f'$ for each function $f$ in Exercises 34-59
$f (x) = \frac{1}{x^{2}}$

Short Answer

Step by step solution

Step 1. Given information

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

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Use the definition of the derivative to find f'for each function fin Exercises 34-59f(x)=1x2

Short Answer

Step by step solution

Step 1. Given information

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

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Mechanics Maths

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Pure Maths

Study anywhere. Anytime. Across all devices.

Use the definition of the derivative to find $f'$ for each function $f$ in Exercises 34-59
$f (x) = \frac{1}{x^{2}}$