Problem 26 - Thomas Calculus

Question

Use the formula $$ f^{\prime}(x)=\lim _{z \rightarrow x} \frac{f(z)-f(x)}{z-x} $$ to find the derivative of the functions in Exercises \(23-26\) $$ g(x)=1+\sqrt{x} $$

Step-by-Step Solution

Answer

The derivative of \( g(x) = 1 + \sqrt{x} \) is \( g'(x) = \frac{1}{2\sqrt{x}} \).

Step 1: Identify the Function

We need to find the derivative of the function given: \( g(x) = 1 + \sqrt{x} \). This means we will apply the formula for the derivative \( g'(x) = \lim_{z \to x} \frac{g(z) - g(x)}{z - x} \).

Step 2: Substitute the Function into the Derivative Formula

Substitute \( g(x) = 1 + \sqrt{x} \) into the derivative formula: \[ g'(x) = \lim_{z \to x} \frac{(1 + \sqrt{z}) - (1 + \sqrt{x})}{z - x} = \lim_{z \to x} \frac{\sqrt{z} - \sqrt{x}}{z - x}. \]

Step 3: Rationalize the Numerator

To simplify \( \frac{\sqrt{z} - \sqrt{x}}{z - x} \), multiply the numerator and the denominator by the conjugate of the numerator: \( \sqrt{z} + \sqrt{x} \):\[ \lim_{z \to x} \frac{(\sqrt{z} - \sqrt{x})(\sqrt{z} + \sqrt{x})}{(z - x)(\sqrt{z} + \sqrt{x})}. \]

Step 4: Simplify the Expression

The numerator becomes:\[ (\sqrt{z} - \sqrt{x})(\sqrt{z} + \sqrt{x}) = z - x. \]Thus the expression now is:\[ \lim_{z \to x} \frac{z - x}{(z - x)(\sqrt{z} + \sqrt{x})} = \lim_{z \to x} \frac{1}{\sqrt{z} + \sqrt{x}}. \]

Step 5: Evaluate the Limit

Now calculate the limit as \( z \to x \):\[ \lim_{z \to x} \frac{1}{\sqrt{z} + \sqrt{x}} = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}. \]

Step 6: Conclude the Derivative

The derivative of the function \( g(x) = 1 + \sqrt{x} \) is:\[ g'(x) = \frac{1}{2\sqrt{x}}. \]

Key Concepts

Rationalizing Expressions

Rationalizing expressions is a method used to remove radicals from the denominators or numerators of algebraic fractions. When dealing with limits in calculus, it often proves beneficial because it simplifies expressions, making them easier to evaluate when approaching a particular point.
To rationalize an expression like \( \frac{\sqrt{z} - \sqrt{x}}{z - x} \), we multiply both the numerator and the denominator by the conjugate \( \sqrt{z} + \sqrt{x} \).

• The conjugate of a binomial of the form \( a - b \) is \( a + b \).
• When you multiply a binomial by its conjugate, you obtain the difference of squares: \( a^2 - b^2 \).

This transforms the original fraction to:
\[ \lim_{z \to x} \frac{(\sqrt{z} - \sqrt{x})(\sqrt{z} + \sqrt{x})}{(z - x)(\sqrt{z} + \sqrt{x})} = \lim_{z \to x} \frac{z - x}{(z - x)(\sqrt{z} + \sqrt{x})}. \]
This simplifies further, as the \( z - x \) terms cancel each other out, simplifying the limit's evaluation.

Limit Definition of Derivative

The limit definition of a derivative is a fundamental concept in calculus. It is written as:
\[ f^{\prime}(x)=\lim _{z \rightarrow x} \frac{f(z)-f(x)}{z-x}. \]
This expression calculates the slope of the tangent line to a function at a particular point \( x \).

• The numerator \( f(z) - f(x) \) represents the change in the function's value as it moves closer to a point.
• The denominator \( z - x \) denotes the change in position.
• The limit \( \lim _{z \rightarrow x} \) ensures the analysis focuses on the behavior as \( z \) approaches \( x \), essentially shrinking the interval to zero.

The goal is to find the derivative, which represents the function's instantaneous rate of change—a key concept when dealing with real-world problems involving rates and speeds.

Function Derivatives

Function derivatives measure how a function changes as its input changes. They provide the rate at which a function is increasing or decreasing at any given point. For the function \( g(x) = 1 + \sqrt{x} \), we want to find its derivative to understand its behavior thoroughly.
The derivative, using the limit definition, involves rationalizing expressions to simplify calculations. After canceling common factors and taking the limit, we reach:
\[ g'(x) = \lim_{z \to x} \frac{1}{\sqrt{z} + \sqrt{x}} = \frac{1}{2\sqrt{x}}. \]

• The final result, \( \frac{1}{2\sqrt{x}} \), tells us how the function's output \( g(x) \) changes with respect to changes in \( x \).
• This derivative will be crucial for determining the slope of the tangent line at any point \( x \) on the curve of \( g(x) \).

Understanding derivatives aids in understanding the behavior of functions, optimizing functions, and solving real-life problems. It's a fundamental tool in calculus and applied mathematics.