Problem 6 - Calculus Concepts

Question

Use the limit definition of the derivative (algebraic method) to confirm the statements. The derivative of \(f(x)=2 x^{0.5}\) is \(f^{\prime}(x)=x^{-0.5}\).

Step-by-Step Solution

Answer

The derivative of \(f(x) = 2x^{0.5}\) is \(f'(x) = x^{-0.5}\).

Step 1: Write the Limit Definition

The limit definition of the derivative for a function \( f(x) \) is given by: \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \). For our function \( f(x) = 2x^{0.5} \), we will substitute \( f(x) \) and \( f(x+h) \) into this formula.

Step 2: Evaluate f(x+h)

Substitute \( x + h \) into the function \( f(x) \): \( f(x+h) = 2(x+h)^{0.5} \). This expression represents the function evaluated at \( x + h \).

Step 3: Set Up the Difference Quotient

Use the expressions for \( f(x+h) \) and \( f(x) \) in the limit definition: \( \frac{f(x+h) - f(x)}{h} = \frac{2(x+h)^{0.5} - 2x^{0.5}}{h} \). This forms the difference quotient.

Step 4: Simplify the Difference Quotient

Rationalize the numerator by multiplying and dividing by its conjugate: \[ \frac{2(x+h)^{0.5} - 2x^{0.5}}{h} \cdot \frac{(x+h)^{0.5} + x^{0.5}}{(x+h)^{0.5} + x^{0.5}} = \frac{2((x+h) - x)}{h((x+h)^{0.5} + x^{0.5})} = \frac{2h}{h((x+h)^{0.5} + x^{0.5})} \].

Step 5: Cancel h in the Quotient

Since \( h eq 0 \), cancel the \( h \) in the numerator and denominator: \( \frac{2}{(x+h)^{0.5} + x^{0.5}} \). This simplification leaves us with an expression that no longer contains \( h \) in the numerator.

Step 6: Evaluate the Limit as h Approaches 0

Finally, compute the limit as \( h \to 0 \): \( \lim_{h \to 0} \frac{2}{(x+h)^{0.5} + x^{0.5}} = \frac{2}{2x^{0.5}} = x^{-0.5} \). This confirms the derivative.

Key Concepts

Limit Definition

To find the derivative of a function, we can start by using the **limit definition** of the derivative. This involves a process of taking small steps to see how the function changes as the input changes slightly. Mathematically, for a function \( f(x) \), the derivative \( f'(x) \) can be defined as:

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

This formula helps us understand how \( f(x) \) changes at a particular point by looking at the average rate of change over an interval that shrinks to zero. We substitute \( f(x) \) into this equation and solve the limit as \( h \) approaches zero. This results in the derivative of the function, giving us a precise measure of its rate of change. In this exercise, we apply this formula to find the derivative of the function \( f(x) = 2x^{0.5} \).

Derivative

The concept of a **derivative** primarily revolves around understanding the rate at which a function changes. When we calculate a derivative, we're essentially finding the slope of the tangent line to the curve at any point \( x \). Consider the function \( f(x) = 2x^{0.5} \). To find its derivative, we'll follow the limit definition.
With \( f'(x) = x^{-0.5} \), this indicates that as \( x \) increases, the function grows at a rate that decreases, which is inverse to the square root of \( x \). Mathematically, the expression \( x^{-0.5} \) can also be written as \( \frac{1}{\sqrt{x}} \), highlighting the inverse relationship. This derivative tells us that small changes in \( x \) yield smaller relative changes in \( f(x) \), demonstrating the decay in growth rate as \( x \) becomes large.

Rationalizing the Numerator

When dealing with limits involving radicals, to simplify the **difference quotient**, we often need to "rationalize the numerator." In this process, we multiply and divide by the conjugate of the numerator, turning square roots into simple expressions. This allows us to simplify the difference and handle limits more easily.

• Original Quotient: \( \frac{2(x+h)^{0.5} - 2x^{0.5}}{h} \)
• To rationalize, we multiply by \( \frac{(x+h)^{0.5} + x^{0.5}}{(x+h)^{0.5} + x^{0.5}} \)
• Result: \( \frac{2((x+h) - x)}{h((x+h)^{0.5} + x^{0.5})} = \frac{2h}{h((x+h)^{0.5} + x^{0.5})} \)

By rationalizing, we simplify the expression and allow terms to cancel, making it easier to evaluate the limit as \( h \to 0 \). This method helps us transform complex radicals into more tractable algebraic expressions.

Difference Quotient

The **difference quotient** is a central part of finding a derivative using the limit definition. It represents the average rate of change of the function \( f(x) \) over an interval \( h \). Technically, for the function \( f(x) = 2x^{0.5} \), the difference quotient is:

• \( \frac{f(x+h) - f(x)}{h} = \frac{2(x+h)^{0.5} - 2x^{0.5}}{h} \)

This expression measures how the function changes as you move from \( x \) to \( x + h \). The next step involves simplifying the difference quotient, which often includes rationalization to eliminate square roots and simplify calculations.
The difference quotient lays the groundwork for taking the limit and ultimately finding the derivative. As \( h \to 0 \), it reveals the instantaneous rate of change, transforming our understanding of the function's behavior from a finite perspective to one at an exact point.