Chapter 2: Q. 47 (page 184)

Use the definition of the derivative to find $f'$ for each function $f$ in Exercises 34-59
$f (x) = 3 \sqrt{x}$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information

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

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

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

putting these values in (1)

$f' (x) = \lim_{h \to 0} \frac{3 \sqrt{x + h} - 3 \sqrt{x}}{h} = \lim_{h \to 0} \frac{3 \sqrt{x + h} - 3 \sqrt{x}}{h} \times \frac{3 \sqrt{x + h} + 3 \sqrt{x}}{3 \sqrt{x + h} + 3 \sqrt{x}} [R a t i o n a l i z i n g] = \lim_{h \to 0} \frac{{(3 \sqrt{x + h})}^{2} - {(3 \sqrt{x})}^{2}}{h [3 (\sqrt{x + h} + \sqrt{x})]} [∵ (a + b) (a - b) = a^{2} - b^{2}] = \lim_{h \to 0} \frac{9 (x + h) - 9 x}{h [3 (\sqrt{x + h} + \sqrt{x})]} = \lim_{h \to 0} \frac{9 x + 9 h - 9 x}{h [3 (\sqrt{x + h} + \sqrt{x})]} = \lim_{h \to 0} \frac{9 h}{3 h [(\sqrt{x + h} + \sqrt{x})]} = \lim_{h \to 0} \frac{3}{(\sqrt{x + h} + \sqrt{x})}$

Putting $h = 0$

$= \frac{3}{\sqrt{x + 0} + \sqrt{x}} = \frac{3}{\sqrt{x} + \sqrt{x}} = \frac{3}{2 \sqrt{x}}$

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

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Use the definition of the derivative to find $f'$ for each function $f$ in Exercises 34-59
$f (x) = 3 \sqrt{x}$

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)=3x

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 34-59
$f (x) = 3 \sqrt{x}$