Problem 58 - Calculus Volume 1

Question

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: Write the Definition of Derivative

The definition of the derivative of a function \(f(x)\) at a point \(x\) is given by the limit: \[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \]

Step 2: Compute \(f(x+h)\)

Substitute \(x+h\) into the function \(f(x) = 5x - x^2\). This gives: \[ f(x+h) = 5(x+h) - (x+h)^2 \] Simplifying, we get: \[ f(x+h) = 5x + 5h - (x^2 + 2xh + h^2) \] Which simplifies further to: \[ f(x+h) = 5x + 5h - x^2 - 2xh - h^2 \]

Step 3: Compute \(f(x+h) - f(x)\)

Subtract \(f(x) = 5x - x^2\) from \(f(x+h)\): \[ f(x+h) - f(x) = (5x + 5h - x^2 - 2xh - h^2) - (5x - x^2) \] Simplify the expression: \[ f(x+h) - f(x) = 5h - 2xh - h^2 \]

Step 4: Form the Difference Quotient

Substitute \(f(x+h) - f(x)\) into the difference quotient: \[ \frac{f(x+h) - f(x)}{h} = \frac{5h - 2xh - h^2}{h} \] Divide each term by \(h\): \[ \frac{f(x+h) - f(x)}{h} = 5 - 2x - h \]

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

Apply the limit as \(h\) approaches 0 to the difference quotient: \[ f'(x) = \lim_{{h \to 0}} (5 - 2x - h) \] Since \(h\) approaches 0, the term \(-h\) vanishes, leaving: \[ f'(x) = 5 - 2x \]

Key Concepts

Definition of Derivative

Understanding the derivative is crucial to mastering calculus. The derivative of a function at a certain point gives the slope of the tangent line at that point. This slope measures how a function changes as its input changes slightly. To find the derivative of a function, we use the definition. For any function \(f(x)\), the derivative \(f'(x)\) at a point \(x\) is defined as the limit:

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

Here, \(h\) represents a very small change in \(x\). This definition captures the idea of the instantaneous rate of change and is fundamental in understanding how functions behave. To calculate a derivative, we often need to manipulate expressions using algebraic skills and then evaluate limits.

Difference Quotient

The difference quotient is a key expression when finding derivatives. It's essentially the ratio of the change in the output to the change in the input of a function. The difference quotient formula is:

• \( \frac{f(x+h) - f(x)}{h} \)

In this formula:- \( f(x+h) \) is the value of the function at a point slightly different from \(x\).- \( f(x) \) is the original value of the function at \(x\).- \( h \) is the small change in the input.By computing this quotient, we are preparing to see how a function's output changes relative to its input. Simplifying the difference quotient is often a preparatory step before applying the limit process. With practice, this calculation becomes straightforward.

Limit Process

To find a derivative using the limit process, we take the simplified difference quotient and formally apply the concept of a limit. The limit, as \(h\) approaches zero, helps us find the precise slope or rate of change at one specific point. For derivatives, the limit transforms an approximate rate of change into an exact value.Let's consider this basic example: if we simplify our difference quotient to a form such as \(5 - 2x - h\), we apply the limit:

• \( f'(x) = \lim_{{h \to 0}} (5 - 2x - h) \)

As \(h\) becomes very tiny and approaches zero, terms involving \(h\) vanish. Therefore, the derivative simplifies to just the constant terms left, indicating the specific slope of the function at any chosen \(x\) point. Understanding this limit process is pivotal, as it allows us to switch from an approximation to precision in mathematical analysis.