Problem 58 - Calculus Volume 1

Question

For the following exercises, use the definition of a derivative to find \(f^{\prime}(x)\) . $$f(x)=5 x-x^{2}$$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 5 - 2x \).

Step 1: Recall the Derivative Definition

The derivative of a function \( f(x) \) at a point \( x \) can be found using the definition: \[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \] This formula represents the limit of the difference quotient as \( h \), the change in \( x \), approaches zero.

Step 2: Substitute the Function into the Definition

First, substitute \( f(x) = 5x - x^2 \) into the definition of the derivative. Start by finding \( f(x + h) \): \[ f(x + h) = 5(x + h) - (x + h)^2 \] Expand this expression: \[ = 5x + 5h - (x^2 + 2xh + h^2) \] \[ = 5x + 5h - x^2 - 2xh - h^2 \]

Step 3: Find the Difference Quotient

Substitute \( f(x + h) \) and \( f(x) \) into the difference quotient: \[ \frac{f(x + h) - f(x)}{h} = \frac{(5x + 5h - x^2 - 2xh - h^2) - (5x - x^2)}{h} \] Simplify the expression: \[ = \frac{5x + 5h - x^2 - 2xh - h^2 - 5x + x^2}{h} \]\[ = \frac{5h - 2xh - h^2}{h} \]

Step 4: Simplify the Fraction

Factor \( h \) out from the numerator: \[ = \frac{h(5 - 2x - h)}{h} \] Cancel \( h \) in the numerator and denominator (provided \( h eq 0 \)): \[ = 5 - 2x - h \]

Step 5: Compute the Limit as \( h \to 0 \)

Now, take the limit as \( h \to 0 \): \[ f'(x) = \lim_{h \to 0} (5 - 2x - h) \] As \( h \to 0 \), the expression simplifies to: \[ = 5 - 2x \] Therefore, the derivative of \( f(x) = 5x - x^2 \) is \( f'(x) = 5 - 2x \).

Key Concepts

Derivative Definition

Understanding the derivative of a function is fundamental in calculus. It helps us know how a function behaves at specific points. The derivative at a point is essentially the slope of the tangent line to the function's graph at that point. It's calculated using a mathematical formula known as the derivative definition.

The formal definition of a derivative for a function \( f(x) \) at a point \( x \) is:

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

What this formula does is measure the rate of change of the function. The difference \( f(x+h) - f(x) \) computes the change in function value as \( x \) changes to \( x + h \). When \( h \) equals zero, the difference quotient gives the slope of the tangent line at the point \( x \).

• **Tangent Line**: A straight line that touches the curve at just one point.
• **Rate of Change**: How fast one quantity changes when another quantity changes.

Difference Quotient

The difference quotient is a crucial component in calculating derivatives. It forms the basis on which the tangent line's slope is determined. In the given context, the difference quotient assesses how much \( f(x) \) changes over a small interval around \( x \).

When \( f(x) = 5x - x^2 \), we need to find \( f(x+h) \), which results in:

• \( f(x+h) = 5(x+h) - (x+h)^2 \)
• Expanding gives \( 5x + 5h - x^2 - 2xh - h^2 \)

Next, substituting this into the difference quotient formula, we have:

• \( \frac{f(x + h) - f(x)}{h} = \frac{5h - 2xh - h^2}{h} \)

This simplifies to \( 5 - 2x - h \), showing the change in \( f(x) \) when divided by the small interval \( h \).

These calculations illustrate how the difference quotient essentially captures the function's rate of change over "tiny" steps, progressing closer to finding the derivative.

Limit Process

Limits are at the heart of defining the derivative. When we compute derivatives, we're mostly concerned with the behavior of a function as some value approaches another value. In the case of the derivative, it’s the limit as \( h \rightarrow 0 \) in the difference quotient that truly defines the derivative.

Here's how the limit plays out in our example:

• The difference quotient we have is \( 5 - 2x - h \).
• As \( h \to 0 \), this expression becomes \( 5 - 2x \).

The limit process allows us to squash the interval \( h \) to zero, effectively turning the average rate of change into an instantaneous rate of change. This resulting value, \( 5 - 2x \), is the derivative of the function, capturing the dynamic behavior at any given point \( x \).

Without considering limits, we wouldn't be able to understand instantaneous change, which is vital when working with continuously varying quantities in calculus. The limit process helps push beyond mere estimation to precisely measure how functions behave instantaneously.