Problem 2 - Calculus an Applied Approach

Question

Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$\begin{array}{ll}{\text { Function }} & {\text { Point }} \\\ {g(x)=(x-4)(x+2)} & {(4,0)} \end{array}$$

Step-by-Step Solution

Answer

The value of the derivative at point (4,0) is 6. The product rule was used to find the derivative.

Step 1: Write down the function

The given function is \(g(x)=(x-4)(x+2)\). We need to break down this function to its basic terms to derive it.

Step 2: Differentiate the function

Since this function is essentially a product of two simpler functions, we can use the product rule for differentiation. The product rule is usually stated as: \((f \cdot g)' = f' \cdot g + f \cdot g'\). Applying this rule to our function, the derivative \(g'(x)\) becomes \((1)*(x+2) + (x-4)*(1) = x + 2 + x - 4 = 2x - 2 \).

Step 3: Evaluate the derivative at the given point

Substitute the x-coordinate from the given point into the derivative. The x-coordinate of the given point (4,0) is 4. Therefore, the derivative at this point is \(g'(4) = 2*4 - 2 = 6\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, which focuses on how a function changes as its input changes. It is the process of finding the derivative of a function, essentially the rate at which one quantity changes with respect to another. In the context of our problem, differentiation helps us understand the rate of change of the function g(x) at any given point on its graph.

g(x) is a simple polynomial and differentiating polynomials is straightforward because they follow well-defined rules, such as the power rule which states that the derivative of x^n is nx^(n-1). Differentiation is critical in fields such as physics, engineering, and economics where the behavior of dynamically changing systems needs to be predicted and analyzed.

Derivatives of Functions

Derivatives of functions represent how a function’s output value changes at any particular point regarding its input value. It gives the slope of the function at any point along the curve. When you calculate a derivative, you’re finding an expression that allows you to find this slope at any point you choose.

In our exercise, the function g(x) is a product of two functions (x - 4) and (x + 2), and we seek to find its derivative to study its slope or rate of change. Derivatives are not just academic; they help predict real-world phenomena, such as velocity in physics, profit changes in economics, and gradients in geography.

Applying the Product Rule

Understanding the Product Rule

When faced with the task of differentiating a function that is the product of two sub-functions, you need the product rule. The product rule states that for two differentiable functions, f(x) and g(x), the derivative of their product f(x)*g(x) is f'(x)g(x) + f(x)g'(x).

In our example, we applied the product rule to differentiate the function g(x)=(x-4)(x+2). We treat each factor as a function: the first function f(x) = (x - 4) and the second g(x) = (x + 2). We then differentiate each and apply the rule to find g'(x). Mastery of the product rule is essential for efficiently dealing with complex functions found in advanced mathematics and applied sciences.

Evaluating Derivatives at a Point

Once you have the derivative of a function, you can evaluate it at any given point to find the instantaneous rate of change of the function at that point. This is done by substituting the input value into the derivative function. In practical terms, evaluating a derivative at a point can tell you the slope of the tangent line to the function’s graph at that specific point.

For the function g(x) = (x - 4)(x + 2), after finding the derivative, we evaluated it at the point (4, 0). Plugging x = 4 into g'(x) = 2x - 2 gives us the value 6, indicating that the slope of the tangent line to the graph of g(x) at x = 4 is 6. This information is crucial for applications involving optimization and modeling physical systems.