Problem 26 - University Calculus: Early Transcendentals

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. $$g(x)=1+\sqrt{x}$$

Step-by-Step Solution

Answer

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

Step 1: Understand the Formula for Derivatives

The formula \( f^{\prime}(x)=\lim _{z \rightarrow x} \frac{f(z)-f(x)}{z-x} \) is used to find the derivative of a function. This requires finding the limit of the expression as \( z \) approaches \( x \).

Step 2: Identify the Function Components

For the function \( g(x) = 1 + \sqrt{x} \), identify \( f(z) \) as \( 1 + \sqrt{z} \) and \( f(x) \) as \( 1 + \sqrt{x} \).

Step 3: Set Up the Difference Quotient

Substitute \( f(z) = 1 + \sqrt{z} \) and \( f(x) = 1 + \sqrt{x} \) into the difference quotient: \[ \frac{f(z) - f(x)}{z - x} = \frac{1 + \sqrt{z} - (1 + \sqrt{x})}{z - x} = \frac{\sqrt{z} - \sqrt{x}}{z - x} \]

Step 4: Simplify the Expression

Simplify the expression \( \frac{\sqrt{z} - \sqrt{x}}{z - x} \). Multiply the numerator and the denominator by the conjugate \( \sqrt{z} + \sqrt{x} \): \[ \frac{\sqrt{z} - \sqrt{x}}{z-x} \cdot \frac{\sqrt{z} + \sqrt{x}}{\sqrt{z} + \sqrt{x}} = \frac{(\sqrt{z} - \sqrt{x})(\sqrt{z} + \sqrt{x})}{(z-x)(\sqrt{z} + \sqrt{x})} \].

Step 5: Evaluate the Limit

The numerator becomes \( \sqrt{z}^2 - \sqrt{x}^2 = z - x \). The expression simplifies to \( \frac{z - x}{(z-x)(\sqrt{z} + \sqrt{x})} = \frac{1}{\sqrt{z} + \sqrt{x}} \). Evaluate the limit as \( z \rightarrow x \): \[ \lim_{z \rightarrow x} \frac{1}{\sqrt{z} + \sqrt{x}} = \frac{1}{2\sqrt{x}} \].

Step 6: Write the Derivative

Thus, the derivative \( g'(x) \) of the function \( g(x) = 1 + \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \).

Key Concepts

Limit Definition of Derivative

The limit definition of a derivative is a foundational concept in calculus that describes how a function changes at a particular point. Understanding this concept involves considering how small changes in the input of a function result in changes in its output. The formula given for this purpose is:
\[f^{\prime}(x)=\lim _{z \rightarrow x} \frac{f(z)-f(x)}{z-x}\]
This formula tells us how to "zoom in" on a function's behavior at a specific point, \(x\). By examining what happens when we make \(z\) very close to \(x\), we can find the rate at which the function \(f\) is changing. This rate is the derivative.Some key takeaways:

• "\(f^{\prime}(x)\)" denotes the derivative of the function at \(x\).
• "\(\lim_{z \rightarrow x}\)" signifies we are looking at what happens as \(z\) gets infinitely close to \(x\).
• The expression \(\frac{f(z) - f(x)}{z - x}\) is what we call a difference quotient.

By using the limit definition, we understand precisely how a function behaves at every little point on its curve, not just overall trends.

Difference Quotient

The difference quotient is an expression that helps in approximating the slope of a function—a crucial step towards understanding derivatives. The form of the difference quotient is:
\[\frac{f(z) - f(x)}{z - x}\]
Think of this as a slope formula. If you've done simple linear equations before, you might recognize it as a version of "rise over run." Here, "rise" is the change in the function's output, \(f(z) - f(x)\), and "run" is the change in input, \(z - x\).Here's what happens during calculus:

• Identify the function's components, \(f(z)\) and \(f(x)\), for substitution into the difference quotient.
• Understand that as \(z\) approaches \(x\), the fraction illustrates how steep the function becomes at \(x\).
• Concluding this process with the limit showcases true function behavior at any point.

This concept is robust because it applies not just to straight lines, but curves which are much more common and complex in mathematics.

Simplifying Expressions

Simplifying expressions is an essential part of solving derivative problems because it transforms complex mathematical expressions into manageable forms. This is particularly useful when dealing with square roots or fractions that might make the derivative look daunting.In our example, when faced with:
\[\frac{\sqrt{z} - \sqrt{x}}{z - x}\]
Multiplying by the conjugate, \(\sqrt{z} + \sqrt{x}\), can eliminate square roots from the numerator, simplifying subsequent calculations.Steps for simplifying the expression:

• Multiply both the numerator and the denominator by the conjugate \(\sqrt{z} + \sqrt{x}\).
• Apply the difference of squares formula to the numerator: \((\sqrt{z} - \sqrt{x})(\sqrt{z} + \sqrt{x}) = z - x\).
• The term \(z - x\) in the numerator cancels with \(z - x\) in the denominator, leaving \(\frac{1}{\sqrt{z} + \sqrt{x}}\).

In calculus, transforming and simplifying expressions is necessary because it allows us to evaluate otherwise difficult limits as straightforward numbers or functions.