4. let ( f(x) = sqrt{2x + 1} ). find ( f(x) ) using the definition of the derivative as a limit. (do not use differentiation formulas.)

5. let ( f(x) = \frac{1}{2x - 1} ). find ( f(x) ) using the definition of the derivative as a limit. (do not use differentiation formulas.)

6. find the derivative using basic differentiation formulas.
(a) ( f(x) = 4x^3 - 3x^2 + x - 1 + \frac{2}{x} + \frac{3}{x^2} )

(b) ( f(x) = 4sqrt{x} - 18sqrt3{x} + \frac{1}{sqrt4{x}} )

Question

4. let ( f(x) = sqrt{2x + 1} ). find ( f(x) ) using the definition of the derivative as a limit. (do not use differentiation formulas.)

5. let ( f(x) = \frac{1}{2x - 1} ). find ( f(x) ) using the definition of the derivative as a limit. (do not use differentiation formulas.)

6. find the derivative using basic differentiation formulas.
(a) ( f(x) = 4x^3 - 3x^2 + x - 1 + \frac{2}{x} + \frac{3}{x^2} )

(b) ( f(x) = 4sqrt{x} - 18sqrt3{x} + \frac{1}{sqrt4{x}} )

Accepted Answer

### Problem 4: Let $ f(x) = \sqrt{x + 1} $. Find $ f'(x) $ using the definition of the derivative as a limit. (Do not use differentiation formulas.)

# Explanation:
## Step 1: Recall the definition of the derivative
The derivative of a function $ f(x) $ is given by:
$$
f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
$$
First, find $ f(x + h) $:
$$
f(x + h) = \sqrt{(x + h) + 1} = \sqrt{x + h + 1}
$$
And $ f(x) = \sqrt{x + 1} $. So the difference quotient is:
$$
\frac{\sqrt{x + h + 1} - \sqrt{x + 1}}{h}
$$

## Step 2: Rationalize the numerator
Multiply the numerator and denominator by the conjugate of the numerator, $ \sqrt{x + h + 1} + \sqrt{x + 1} $:
$$
\frac{(\sqrt{x + h + 1} - \sqrt{x + 1})(\sqrt{x + h + 1} + \sqrt{x + 1})}{h(\sqrt{x + h + 1} + \sqrt{x + 1})}
$$
The numerator is a difference of squares: $ (a - b)(a + b) = a^2 - b^2 $, so:
$$
\frac{(x + h + 1) - (x + 1)}{h(\sqrt{x + h + 1} + \sqrt{x + 1})}
$$

## Step 3: Simplify the numerator
Simplify the numerator:
$$
\frac{x + h + 1 - x - 1}{h(\sqrt{x + h + 1} + \sqrt{x + 1})} = \frac{h}{h(\sqrt{x + h + 1} + \sqrt{x + 1})}
$$
Cancel the $ h $ in the numerator and denominator (assuming $ h
eq 0 $, which is valid since we are taking the limit as $ h \to 0 $):
$$
\frac{1}{\sqrt{x + h + 1} + \sqrt{x + 1}}
$$

## Step 4: Take the limit as $ h \to 0 $
Now, take the limit as $ h \to 0 $:
$$
\lim_{h \to 0} \frac{1}{\sqrt{x + h + 1} + \sqrt{x + 1}}
$$
As $ h \to 0 $, $ \sqrt{x + h + 1} \to \sqrt{x + 1} $, so:
$$
\frac{1}{\sqrt{x + 1} + \sqrt{x + 1}} = \frac{1}{2\sqrt{x + 1}}
$$

# Answer:
$ f'(x) = \frac{1}{2\sqrt{x + 1}} $

### Problem 5: Let $ f(x) = \frac{1}{2x - 1} $. Find $ f'(x) $ using the definition of the derivative as a limit. (Do not use differentiation formulas.)

# Explanation:
## Step 1: Recall the definition of the derivative
The derivative of a function $ f(x) $ is given by:
$$
f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
$$
First, find $ f(x + h) $:
$$
f(x + h) = \frac{1}{2(x + h) - 1} = \frac{1}{2x + 2h - 1}
$$
And $ f(x) = \frac{1}{2x - 1} $. So the difference quotient is:
$$
\frac{\frac{1}{2x + 2h - 1} - \frac{1}{2x - 1}}{h}
$$

## Step 2: Combine the fractions in the numerator
Find a common denominator for the numerator: $ (2x + 2h - 1)(2x - 1) $
$$
\frac{(2x - 1) - (2x + 2h - 1)}{(2x + 2h - 1)(2x - 1)h}
$$

## Step 3: Simplify the numerator
Simplify the numerator:
$$
\frac{2x - 1 - 2x - 2h + 1}{(2x + 2h - 1)(2x - 1)h} = \frac{-2h}{(2x + 2h - 1)(2x - 1)h}
$$
Cancel the $ h $ in the numerator and denominator (assuming $ h
eq 0 $):
$$
\frac{-2}{(2x + 2h - 1)(2x - 1)}
$$

## Step 4: Take the limit as $ h \to 0 $
Now, take the limit as $ h \to 0 $:
$$
\lim_{h \to 0} \frac{-2}{(2x + 2h - 1)(2x - 1)}
$$
As $ h \to 0 $, $ 2x + 2h - 1 \to 2x - 1 $, so:
$$
\frac{-2}{(2x - 1)(2x - 1)} = \frac{-2}{(2x - 1)^2}
$$

# Answer:
$ f'(x) = \frac{-2}{(2x - 1)^2} $

### Problem 6(a): Find the derivative of $ f(x) = 4x^3 - 3x^2 + x - 1 + \frac{2}{x} + \frac{3}{x^2} $ using basic differentiation formulas.

# Explanation:
## Step 1: Rewrite the terms with negative exponents
Rewrite $ \frac{2}{x} $ as $ 2x^{-1} $ and $ \frac{3}{x^2} $ as $ 3x^{-2} $:
$$
f(x) = 4x^3 - 3x^2 + x - 1 + 2x^{-1} + 3x^{-2}
$$

## Step 2: Apply the power rule and sum/difference rule
The power rule is $ \frac{d}{dx}(x^n) = nx^{n - 1} $, and the derivative of a constant is 0.
- Derivative of $ 4x^3 $: $ 4 \cdot 3x^{3 - 1} = 12x^2 $
- Derivative of $ -3x^2 $: $ -3 \cdot 2x^{2 - 1} = -6x $
- Derivative of $ x $: $ 1 \cdot x^{1 - 1} = 1 $
- Derivative of $ -1 $: $ 0 $
- Derivative of $ 2x^{-1} $: $ 2 \cdot (-1)x^{-1 - 1} = -2x^{-2} $
- Derivative of $ 3x^{-2} $: $ 3 \cdot (-2)x^{-2 - 1} = -6x^{-3} $

## Step 3: Combine the derivatives
Combine the results:
$$
f'(x) = 12x^2 - 6x + 1 - 2x^{-2} - 6x^{-3}
$$
Rewrite back with positive exponents for the last two terms:
$$
f'(x) = 12x^2 - 6x + 1 - \frac{2}{x^2} - \frac{6}{x^3}
$$

# Answer:
$ f'(x) = 12x^2 - 6x + 1 - \frac{2}{x^2} - \frac{6}{x^3} $

### Problem 6(b): Find the derivative of $ f(x) = 4\sqrt{x} - 18\sqrt[3]{x} + \frac{1}{\sqrt[4]{x}} $ using basic differentiation formulas.

# Explanation:
## Step 1: Rewrite the terms with fractional exponents
Rewrite $ \sqrt{x} $ as $ x^{\frac{1}{2}} $, $ \sqrt[3]{x} $ as $ x^{\frac{1}{3}} $, and $ \frac{1}{\sqrt[4]{x}} $ as $ x^{-\frac{1}{4}} $:
$$
f(x) = 4x^{\frac{1}{2}} - 18x^{\frac{1}{3}} + x^{-\frac{1}{4}}
$$

## Step 2: Apply the power rule
The power rule is $ \frac{d}{dx}(x^n) = nx^{n - 1} $.
- Derivative of $ 4x^{\frac{1}{2}} $: $ 4 \cdot \frac{1}{2}x^{\frac{1}{2} - 1} = 2x^{-\frac{1}{2}} $
- Derivative of $ -18x^{\frac{1}{3}} $: $ -18 \cdot \frac{1}{3}x^{\frac{1}{3} - 1} = -6x^{-\frac{2}{3}} $
- Derivative of $ x^{-\frac{1}{4}} $: $ -\frac{1}{4}x^{-\frac{1}{4} - 1} = -\frac{1}{4}x^{-\frac{5}{4}} $

## Step 3: Rewrite back with radicals
Rewrite the terms with negative exponents as radicals:
- $ 2x^{-\frac{1}{2}} = \frac{2}{\sqrt{x}} $
- $ -6x^{-\frac{2}{3}} = -\frac{6}{\sqrt[3]{x^2}} $
- $ -\frac{1}{4}x^{-\frac{5}{4}} = -\frac{1}{4x^{\frac{5}{4}}} = -\frac{1}{4\sqrt[4]{x^5}} = -\frac{1}{4x\sqrt[4]{x}} $ (optional simplification)

Combining these, we get:
$$
f'(x) = \frac{2}{\sqrt{x}} - \frac{6}{\sqrt[3]{x^2}} - \frac{1}{4x^{\frac{5}{4}}}
$$

# Answer:
$ f'(x) = \frac{2}{\sqrt{x}} - \frac{6}{\sqrt[3]{x^2}} - \frac{1}{4x^{\frac{5}{4}}} $ (or simplified further as $ \frac{2}{\sqrt{x}} - \frac{6}{\sqrt[3]{x^2}} - \frac{1}{4x\sqrt[4]{x}} $)

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QUESTION IMAGE

Question

Explanation:

Problem 4: Let \( f(x) = \sqrt{x + 1} \). Find \( f'(x) \) using the definition of the derivative as a limit. (Do not use differentiation formulas.)

Step 1: Recall the definition of the derivative

Step 2: Rationalize the numerator

Step 3: Simplify the numerator

Step 4: Take the limit as \( h \to 0 \)

Step 1: Recall the definition of the derivative

Step 2: Combine the fractions in the numerator

Step 3: Simplify the numerator

Step 4: Take the limit as \( h \to 0 \)

Step 1: Rewrite the terms with negative exponents

Step 2: Apply the power rule and sum/difference rule

Step 3: Combine the derivatives

Answer:

Problem 5: Let \( f(x) = \frac{1}{2x - 1} \). Find \( f'(x) \) using the definition of the derivative as a limit. (Do not use differentiation formulas.)