Question

Finding and Evaluating a Derivative In Exercises \(13-18,\) find \(f^{\prime}(x)\) and \(f^{\prime}(c) .\) $$ \text{Function} \quad \text {Value of} \quad c $$ $$ f(x)=\frac{x^{2}-4}{x-3} \quad c=1 $$

Short answer

The derivative of the function \(f(x)= \frac{x^{2}-4}{x-3}\) is \(f^{\prime}(x) = \frac{x^{2} - 6x + 4}{(x - 3)^2}\) and the value of the derivative at the point \(x = c\) is \(f^{\prime}(c) = -\frac{1}{4}\)

Step-by-step solution

Differentiate the Function

The function in this case is a quotient, we need to use the quotient rule for derivative which is \((f/g)' = (f'g - fg') / g^2\). In our case, \(f(x) = x^2 - 4\) and \(g(x) = x - 3\). Hence, \(f'(x) = 2x\) and \(g'(x) = 1\). Then, the derivative of the function \(f(x)\) will be \[f^{\prime}(x)=\frac{(2x \cdot (x-3) - ((x^2 - 4) \cdot 1))}{(x-3)^2}\] which simplifies to \[f^{\prime}(x) = \frac{2x^{2} - 6x - x^{2} + 4}{(x - 3)^2} = \frac{x^{2} - 6x + 4}{(x - 3)^2}\]

Find the Value of the Derivative at the Point x = c

Once the derivative is found, it can now be evaluated at \(x = c\). Substitute 1 for \(x\) in the derivative \(f^{\prime}(x)\) gives \[f^{\prime}(1) = \frac{(1)^{2} - 6 * 1 + 4}{(1 - 3)^2}\] which simplifies to \[f^{\prime}(c) = -\frac{1}{4}\]

Key concepts

Derivative

A derivative is a fundamental concept in calculus that represents the rate of change of a function with respect to a variable. It's analogous to how fast something is changing in terms of its position, speed, or other attributes. In many ways, finding a derivative is akin to finding the slope of the tangent line to the curve at a particular point. If you think of a function as a journey, the derivative can be thought of as the speedometer. It tells you how fast you are traveling at any given point along that journey. For example, if you have a function that represents the position of a car over time, its derivative at any given point will tell you the car's speed at that instant in time. Derivatives are used widely in various fields such as physics for motion analysis, in economics for finding optimal points, and in biology for modeling population growth, among others. In the context of our exercise, the derivative of the function gives us insight into how the function behaves around the point where we evaluate the derivative at a given value of `c`.

Quotient Rule

When differentiating a function that is the quotient of two other functions, like in this exercise, one must use the quotient rule. The quotient rule is specifically designed to handle the derivative of a division of two functions. It can be written as: \((\frac{f}{g})' = \frac{f'g - fg'}{g^2}\)Here, `f` and `g` are functions whose derivatives are `f'` and `g'`, respectively.

• The numerator of the rule formula involves multiplying the derivative of the numerator part by the denominator part, and then subtracting the product of the numerator and the derivative of the denominator.
• The denominator of the rule formula is simply the square of the original denominator.

In our example, the functions are \(f(x) = x^2 - 4\) for the numerator and \(g(x) = x - 3\) for the denominator. Following the quotient rule carefully allows us to differentiate these functions with ease, leading us to the resultant derivative, \(f'(x)\). This rule is essential for differentiating quotients because it accounts for the interdependent variability between the numerator and the denominator.

Differentiation

Differentiation refers to the process of computing a derivative. It's the method used to find out how a function changes and behaves. Whether you're dealing with simple linear functions or complex logarithmic or trigonometric ones, differentiation helps in understanding these changes concretely.In this exercise, differentiation involves applying the quotient rule to the function \(f(x) = \frac{x^2 - 4}{x - 3}\). By identifying the parts that make up the quotient (numerator and denominator), we first differentiate each of these parts separately, before using these results to apply the quotient rule.Choosing the right strategy for differentiation depends on the form of the function. Using the rule that best suits the function's structure ensures accuracy and efficiency in finding the derivative. Practicing differentiation across different function forms strengthens mathematical problem-solving skills and aids in recognizing patterns across problems.

Evaluating Derivatives

Once a derivative is obtained, the next step is often to evaluate it at a specific point. This means substituting a designated value into the derivative to assess the rate of change at that point. In this context, `c` signifies a specific point where the derivative of the function \(f(x)\) is to be evaluated.For the function we differentiated, \(f'(x) = \frac{x^{2} - 6x + 4}{(x - 3)^2}\), evaluating the derivative at \(x = 1\) involves substituting \(1\) for \(x\) in this expression, resulting in:\[f^{\prime}(1) = \frac{1^{2} - 6 \times 1 + 4}{(1 - 3)^2}\]Simplifying the expression yields \(f'(1) = -\frac{1}{4}\).

• This evaluation tells us how the changes in \(f(x)\) are behaving precisely at \(x = 1\).
• It gives a local perspective of the function's behavior, showing the slope of the tangent line to the curve at that point.

Evaluating derivatives at specific points is crucial for understanding real-world applications, as it provides specific insights into how things change at particular moments, assisting in decision-making and predictions.