Question

Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$ g(x)=(x-4)(x+2) $$

Short answer

The derivative of the function \(g(x) = (x - 4)(x + 2)\) is \(g'(x) = 2x - 2\). This derivative can then be evaluated at the specific point (x-value) to obtain the slope at that point on the graph of the function.

Step-by-step solution

Understanding the Product Rule of Differentiation

The product rule of differentiation states that the derivative of the product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function. Given in mathematical terms \[ (f(x)*g(x))' = f'(x)*g(x) + f(x)*g'(x) \]

Identify the Functions and Apply the Product Rule

Here, \(f(x) = x - 4\) and \(g(x) = x + 2\). Now apply the product rule: \[g'(x) = f'(x)*g(x) + f(x)*g'(x) = (1)*(x + 2) + (x - 4)*(1) = x + 2 + x - 4 = 2x - 2 \]

Find the Derivative at the Given Point

The problem however, asks for the derivative at a specific point which has not been provided. The student would evaluate the derivative \(2x - 2\) at this point.

Key concepts

Derivative Calculation

Understanding how to compute derivatives is an essential part of calculus, a field dealing primarily with change. When students first learn about the product rule of differentiation, they are grasping a technique for handling the rate of change in scenarios where two varying quantities—each affecting the other—are multiplied together.

Let’s imagine that you're trying to find out how quickly a rectangular area expands when both the length and width increase. In calculus, were you to write the area as the product of two functions—for instance, length and width—you would apply the product rule to find the rate at which the area changes with respect to time.

The product rule formula, \[ (f(x) * g(x))' = f'(x) * g(x) + f(x) * g'(x) \], is applied when both factors of the product are functions of the same variable and are differentiable. Remember that the derivative, symbolized by the prime mark \(f'(x)\text{or}g'(x)\), represents the slope of the tangent line to the function at any point, indicating how quickly the function value changes as its input changes.

Calculus

Calculus is a vast field of mathematics that provides tools for modeling and analyzing dynamic systems. It encompasses both differential calculus (concerned with instantaneous rates of change and slopes of curves) and integral calculus (focused on accumulation of quantities and areas under or between curves).

When faced with a function like \(g(x) = (x - 4)(x + 2)\), a student will use calculus concepts, particularly derivative rules like the product rule, to understand how the function behaves as input values change. In practical terms, calculus helps predict outcomes in physics, economics, engineering, and beyond.

A clear understanding of derivative calculation allows one to smoothly transition to more complex applications such as optimization problems where one might want to find maximum profits or minimum cost of production in a business setting, or maximum efficiency in an engineering project.

Applied Algebra

Applied algebra plays a pivotal role in calculus, particularly in simplifying expressions and solving equations to make derivative calculations more manageable. For instance, in the product rule application to \(g(x) = (x - 4)(x + 2)\), algebra is used to simplify the derivatives \(f'(x)\) and \(g'(x)\) to their basic forms before applying the product rule.

The simplification process often looks something like this: take \(f(x) = x - 4\) and \(g(x) = x + 2\). The derivatives are \(f'(x) = 1\) and \(g'(x) = 1\), obtained by knowing that the derivative of \(x\) is \(1\) and the derivative of a constant is \(0\). Post simplification, applying the product rule involves basic algebraic steps to combine like terms and simplify the result, ultimately revealing how algebra serves as the scaffolding for calculus problems.