Chapter 2: Q. 29 (page 184)

Use (a) the $h \to 0$ definition of the derivative and then
(b) the $z \to c$ definition of the derivative to find $f' (c)$ for each function f and value $x = c$ in Exercises 23–38.
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

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

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

We have

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

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Use (a) the $h \to 0$ definition of the derivative and then
(b) the $z \to c$ definition of the derivative to find $f' (c)$ for each function f and value $x = c$ in Exercises 23–38.
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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Use (a) the h→0definition of the derivative and then(b) the z→cdefinition of the derivative to find f'(c) for each function f and value x=c in Exercises 23–38.29.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

Theoretical and Mathematical Physics

Calculus

Pure Maths

Geometry

Applied Mathematics

Logic and Functions

Study anywhere. Anytime. Across all devices.

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