Q. 50

Question

Use the definition of the derivative to find $f'$ for each function $f$ in Exercises 39-54

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

Step-by-Step Solution

Verified

Answer

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

1Step 1. Given information

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

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

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

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

Q. 49

Q. 50

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Q. 50

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