Problem 63 - Calculus Volume 1

Question

For the following exercises, use the definition of a derivative to find \(f^{\prime}(x)\) . $$f(x)=\frac{1}{\sqrt{x}}$$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{-1}{2x\sqrt{x}} \).

Step 1: Understand the Definition of a Derivative

The derivative of a function \(f(x)\) at a point \(x=a\) is given by the limit \(f'(x)= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\). We will use this definition to find the derivative of \(f(x) = \frac{1}{\sqrt{x}}\).

Step 2: Setup the Derivative Formula

First, we will substitute \(f(x)\) into the derivative formula. We have \(f(x+h) = \frac{1}{\sqrt{x+h}}\), so the expression becomes \(f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h}\).

Step 3: Simplify the Expression

To simplify \(\frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h}\), we'll multiply the numerator and the denominator by the conjugate of the numerator: \(\frac{1}{\sqrt{x+h}} + \frac{1}{\sqrt{x}}\).

Step 4: Multiply by Conjugate

Apply \(\frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}\) to get \(\frac{(\sqrt{x} - \sqrt{x+h})}{h(\sqrt{x+h} + \sqrt{x})}\). This simplifies to \(\frac{h}{h(\sqrt{x+h}\sqrt{x})}\).

Step 5: Cancel and Simplify Further

Cancel \(h\) in the numerator and denominator: \(\lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt{x})(\sqrt{x+h} + \sqrt{x})}\).

Step 6: Evaluate the Limit

As \(h\to0\), \(\sqrt{x+h}\to \sqrt{x}\). Therefore: \(f'(x) = \frac{-1}{2x\sqrt{x}}\).

Step 7: Final Derivative Expression

So the derivative \(f'(x)\) is: \[ f'(x) = \frac{-1}{2x\sqrt{x}} \]

Key Concepts

Limit Definition of Derivative

The limit definition of a derivative is a fundamental concept in calculus. It provides a way to precisely calculate the rate at which a function is changing at any given point. For a function \( f(x) \), the derivative is expressed as:

• \( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \)

This formula effectively measures the instantaneous rate of change or slope of the function's graph at a particular point.
In our exercise, \( f(x) = \frac{1}{\sqrt{x}} \), we substitute the function into the formula to set up:
\( f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h} \).

This is the foundational step in finding the derivative and involves understanding how the limit concept helps in defining derivatives.

Simplifying Rational Expressions

Simplifying rational expressions involves breaking down complex fractions into simpler forms. This process is essential when finding derivatives using limits as it makes handling complex expressions manageable.

• In the derivative expression \( \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h} \), the two terms in the numerator form a complex rational expression.

By simplifying it, we aim to make it simpler for further mathematical operations like applying limits.
To start simplifying, note that we plan to remove the fractions within a fraction by combining them into a single rational expression.
This process often entails rationalizing or using algebraic manipulation techniques like finding a common denominator.

Conjugate Multiplication

Conjugate multiplication is a powerful algebraic technique used to simplify expressions involving square roots. It involves multiplying by a conjugate, which eliminates square roots in the numerator or denominator.

• The conjugate of \( \frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}} \) is \( \frac{1}{\sqrt{x+h}} + \frac{1}{\sqrt{x}} \).

By multiplying the expression by its conjugate, we get:
\( \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} \).

Performing this multiplication simplifies complex square root expressions by applying the difference of squares formula, which reduces the expression to a manageable form before evaluating limits.

Evaluating Limits

Evaluating limits is the final step in finding the derivative using the limit definition. After simplifying expressions, the substitution \( h \to 0 \) allows us to find the exact value of the derivative.
Upon simplification, we obtained:
\( \frac{h}{h(\sqrt{x+h}\sqrt{x})} \).
Cancel the \( h \) in numerator and denominator, resulting in:
\( \lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt{x})(\sqrt{x+h} + \sqrt{x})} \).
This simplification depends on knowing that as \( h \to 0 \), \( \sqrt{x+h} \to \sqrt{x} \).

We find that \( f'(x) = \frac{-1}{2x\sqrt{x}} \).
Evaluating limits ensures all algebraic manipulations lead us to the correct result.