Question

Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$ g(x)=\left(x^{2}-4 x+3\right)(x-2) $$

Short answer

The derivative of the function \(g(x)\) is \(g'(x)=3x^2-12x+3\). To evaluate the derivative at a given point \(x=a\), replace \(x\) with \(a\) in the derivative expression.

Step-by-step solution

Identification of the Function Form

The given function \( g(x)=(x^2-4x+3)(x-2) \) is in the form of \(u(x)v(x)\) which requires the use of product rule for differentiation.

Application of the Product Rule

The product rule is given by: \((u.v)' = u'v + uv'\). First, let \(u(x)=x^2-4x+3\) and \(v(x)=x-2\). Then, differentiate \(u(x)\) and \(v(x)\) to get \(u'(x)\) and \(v'(x)\) respectively. We get \(u'(x)=2x-4\) and \(v'(x)=1\).

Substituting the values

Substitute \(u(x)\), \(v(x)\), \(u'(x)\), and \(v'(x)\) into the product rule formula. Hence, \((uv)'= (2x-4)(x-2) + (x^2-4x+3)(1)\). That simplifies to \( g'(x)=2x^2-8x+x^2-4x+3=3x^2-12x+3\).

Evaluate the Derivative at the Given Point

The point of evaluation is not given in the problem. But, if it were given, for instance as \(x=a\), then simply substitute \(x=a\) in the obtained derivative expression.

Key concepts

Derivative of a Function

The derivative of a function represents the rate at which the function's value is changing at a certain point. In simpler terms, think of it as how quickly the function's output is moving up or down as you move along the function's graph.

For example, if you're driving a car and the function represents your distance over time, the derivative is your speed at each moment. In calculus, you'll often see the derivative notated as \( f'(x) \) or \( \frac{df}{dx} \). When you take the derivative of a function, you're finding a new function that gives you the slope of the original function at any point.

Understanding derivatives is crucial as they form the basis for a significant portion of calculus, including optimization problems and understanding motion.

Differentiation Rules

Differentiation is full of rules that make the process of finding derivatives systematic and predictable. One of the most vital rules is the product rule, which is used when you need to differentiate the product of two functions. The formula is \( (uv)' = u'v + uv' \), where \( u \) and \( v \) are functions of \( x \).

Other key rules include the power rule, which is used for differentiating powers of \( x \), and the sum rule, that allows you to differentiate expressions term by term. The quotient rule and chain rule are also essential differentiation rules for dividing functions and for compositions of functions, respectively. Each rule has a specific use case, helping simplify complex differentiation problems into more manageable pieces.

Applied Calculus

Applied calculus is the application of calculus to solve problems in fields such as science, engineering, economics, and business. It usually involves setting up a mathematical model and using calculus to find an optimal solution. For instance, you might use derivatives to find the maximum profit for a business or the least time for an object to reach a certain point.

In our textbook example, applied calculus can be used to determine the maximum or minimum points of the function \( g(x) \) or how fast \( g(x) \) is increasing or decreasing at a certain value of \( x \). These applications of calculus are essential for real-world problem-solving and decision-making.

Algebraic Functions

Algebraic functions are functions that can be constructed using algebraic operations—addition, subtraction, multiplication, division, and taking roots—on polynomial functions. The given function in the exercise, \( g(x)=(x^2-4x+3)(x-2) \), is an example of an algebraic function because it's formed by multiplying two polynomials.

These functions often form the basis of more complex calculus problems. Understanding how to manipulate these functions algebraically is essential before applying calculus rules to them as seen by the necessity of differentiating \( u(x) \) and \( v(x) \) before applying the product rule. Mastering both algebra and calculus is crucial for tackling advanced mathematical problems in a variety of fields.