Chapter 10: Q. 43 (page 777)

Use the definition of the derivative to find $f'$ for each function $f$ in Exercises 39-54
$f (x) = \frac{2}{x + 1}$

Short Answer

Expert verified

The value of $f' (x) = \frac{- 2}{{(x + 1)}^{2}}$

Step by step solution

Step 1. Given information

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

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 $f (x) = \frac{2}{x + 1}$ then

$f (x + h) = \frac{2}{(x + h) + 1}$

Putting these values in (1)

$f' (x) = \lim_{h \to 0} \frac{\frac{2}{(x + h) + 1} - \frac{2}{x + 1}}{h} = \lim_{h \to 0} \frac{\frac{2 (x + 1) - 2 (x + h + 1)}{(x + h + 1) (x + 1)}}{h} = \lim_{h \to 0} \frac{\frac{2 x + 2 - 2 x - 2 h - 2}{(x + h + 1) (x + 1)}}{h} = \lim_{h \to 0} \frac{\frac{- 2 h}{(x + h + 1) (x + 1)}}{h} = \lim_{h \to 0} \frac{- 2 h}{(x + h + 1) (x + 1)} \times \frac{1}{h} = \lim_{h \to 0} \frac{- 2}{(x + h + 1) (x + 1)}$

Putting $h = 0$

$= \frac{- 2}{(x + 0 + 1) (x + 1)} = \frac{- 2}{(x + 1) (x + 1)} = \frac{- 2}{{(x + 1)}^{2}}$

Hence, the value of $f' (x) = \frac{- 2}{{(x + 1)}^{2}}$

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Use the definition of the derivative to find $f'$ for each function $f$ in Exercises 39-54
$f (x) = \frac{2}{x + 1}$

Short Answer

Step by step solution

Step 1. Given information

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

One App. One Place for Learning.

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Use the definition of the derivative to find f'for each function fin Exercises 39-54f(x)=2x+1

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

Discrete Mathematics

Mechanics Maths

Applied Mathematics

Pure Maths

Theoretical and Mathematical Physics

Statistics

Study anywhere. Anytime. Across all devices.

Use the definition of the derivative to find $f'$ for each function $f$ in Exercises 39-54
$f (x) = \frac{2}{x + 1}$