Chapter 10: Q. 50 (page 777)

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

Short Answer

Expert verified

The value of $f' (x) = - {(2 x + 1)}^{- \frac{3}{2}}$

Step by step solution

Step 1. Given information

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

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

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

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

Then role="math" localid="1648288671514" $f (x + h) = \frac{1}{\sqrt{2 a + 2 h + 1}}$

$f' (x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{2 x + 2 h + 1}} - \frac{1}{\sqrt{2 x + 1}}}{h} = \lim_{h \to 0} \frac{\sqrt{2 x + 1} - \sqrt{2 x + 2 h + 1}}{h \sqrt{2 x + 2 h + 1} \sqrt{2 x + 1}} = \lim_{h \to 0} \frac{(\sqrt{2 x + 1} - \sqrt{2 x + 2 h + 1}) (\sqrt{2 x + 1} + \sqrt{2 x + 2 h + 1)}}{h \sqrt{2 x + 2 h + 1} \sqrt{2 a + 1} (\sqrt{2 x + 1} + \sqrt{2 x + 2 h + 1})} = \lim_{h \to 0} \frac{(2 x + 1) - (2 x + 2 h + 1)}{h \sqrt{2 x + 2 h + 1} \sqrt{2 x + 1} (\sqrt{2 x + 1} + \sqrt{2 x + 2 h + 1})} = \lim_{h \to 0} \frac{- 2 h}{h \sqrt{2 x + 2 h + 1} \sqrt{2 x + 1} (\sqrt{2 x + 1} + \sqrt{2 x + 2 h + 1})} P u t t i n g h = 0 = \frac{- 2}{\sqrt{2 x + 1} \sqrt{2 x + 1} (\sqrt{2 x + 1} + \sqrt{2 x + 1})} = \frac{- 1}{{(2 x + 1)}^{\frac{3}{2}}} = - {(2 x + 1)}^{- \frac{3}{2}}$

Hence, the value of $f' (x) = - {(2 x + 1)}^{- \frac{3}{2}}$

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

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 39-54f(x)=12x+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

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