Q. 29 Use (a) the h鈫�0聽definition of... [FREE SOLUTION]

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Chapter 2: Q. 29 (page 184)

Use (a) the $h 鈫� 0$ definition of the derivative and then
(b) the $z 鈫� c$ definition of the derivative to find $f' (c)$ for each function f and value $x = c$ in Exercises 23鈥�38.
29. $f (x) = x^{\frac{1}{2}}, x = 9$

Short Answer

Expert verified

$f' (9) = \frac{1}{6}$ .

Step by step solution

Part (a) Step 1. Given information.

A function is given as $f (x) = x^{\frac{1}{2}}$ and $x = c = 9$ .

Part (a) Step 2. Solve f'(9) using h→0 definition of the derivative.

We have

$f' (c) = \lim_{h 鈫� 0} \frac{f (c + h) - f (c)}{h} f' (9) = \lim_{h 鈫� 0} \frac{f (9 + h) - f (9)}{h} = \lim_{h 鈫� 0} \frac{{(9 + h)}^{\frac{1}{2}} - 9^{\frac{1}{2}}}{h} = \lim_{h 鈫� 0} \frac{\sqrt{9 + h} - \sqrt{9}}{h} = \lim_{h 鈫� 0} \frac{\sqrt{9 + h} - \sqrt{9}}{h} 脳 \frac{\sqrt{9 + h} + \sqrt{9}}{\sqrt{9 + h} + \sqrt{9}} = \lim_{h 鈫� 0} \frac{9 + h - 9}{h (\sqrt{9 + h} + \sqrt{9})} = \lim_{h 鈫� 0} \frac{h}{h (\sqrt{9 + h} + \sqrt{9})} = \lim_{h 鈫� 0} \frac{1}{\sqrt{9 + h} + \sqrt{9}} = \frac{1}{\sqrt{9 + 0} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

Part (b) Step 1. Solve f'(9) using z→9 definition of the derivative.

We have

$f' (c) = \lim_{z 鈫� c} \frac{f (z) - f (c)}{z - c} f' (9) = \lim_{z 鈫� 9} \frac{f (z) - f (9)}{z - 9} = \lim_{z 鈫� 9} \frac{\sqrt{z} - \sqrt{9}}{z - 9} = \lim_{z 鈫� 9} \frac{\sqrt{z} - \sqrt{9}}{{(\sqrt{z})}^{2} - {(\sqrt{9})}^{2}} = \lim_{z 鈫� 9} \frac{\sqrt{z} - \sqrt{9}}{(\sqrt{z} - \sqrt{9}) (\sqrt{z} + \sqrt{9})} = \lim_{z 鈫� 9} \frac{1}{\sqrt{z} + \sqrt{9}} = \lim_{z 鈫� 9} \frac{1}{\sqrt{z} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

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91影视

Use (a) the $h 鈫� 0$ definition of the derivative and then
(b) the $z 鈫� c$ definition of the derivative to find $f' (c)$ for each function f and value $x = c$ in Exercises 23鈥�38.
29. $f (x) = x^{\frac{1}{2}}, x = 9$

Short Answer

Step by step solution

Part (a) Step 1. Given information.

Part (a) Step 2. Solve f'(9) using h→0 definition of the derivative.

Part (b) Step 1. Solve f'(9) using z→9 definition of the derivative.

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91影视

Use (a) the h鈫�0definition of the derivative and then(b) the z鈫�cdefinition of the derivative to find f'(c) for each function f and value x=c in Exercises 23鈥�38.29.f(x)=x12,x=9

Short Answer

Step by step solution

Part (a) Step 1. Given information.

Part (a) Step 2. Solve f'(9) using h→0 definition of the derivative.

Part (b) Step 1. Solve f'(9) using z→9 definition of the derivative.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Discrete Mathematics

Calculus

Logic and Functions

Pure Maths

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Use (a) the $h 鈫� 0$ definition of the derivative and then
(b) the $z 鈫� c$ definition of the derivative to find $f' (c)$ for each function f and value $x = c$ in Exercises 23鈥�38.
29. $f (x) = x^{\frac{1}{2}}, x = 9$