Question

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

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

1. 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}} )

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

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:

$$ \frac{1}{\sqrt{x + 1} + \sqrt{x + 1}} = \frac{1}{2\sqrt{x + 1}} $$

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:

$$ \frac{-2}{(2x - 1)(2x - 1)} = \frac{-2}{(2x - 1)^2} $$

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:

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

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.)