Problem 13 - College Algebra and 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. $$ g(x)=\frac{2 x+1}{x-5} $$

Step-by-Step Solution

Answer

The derivative of the function g(x) is given by g'(x) = -6/(x-5)^2, obtained using the Quotient Rule. To find its value at a specific point, you would need to substitute that point value into this expression.

Step 1: State the Quotient Rule

The Quotient Rule in differentiation states that for two functions u and v, the derivative of their quotient, (u/v)', is given by (u' * v - u * v') / (v^2). In our case, u(x) = 2x + 1 and v(x) = x - 5.

Step 2: Find Derivatives of U and V

The derivative of u(x) = 2x + 1 with respect to x is u'(x) = 2. The derivative of v(x) = x - 5 with respect to x is v'(x) = 1.

Step 3: Apply the Quotient Rule

Substitute u, v, u' and v' into the Quotient Rule. The derivative of g(x), g'(x), is (2*(x-5) - (2x+1)*1) / ((x-5)^2). That simplifies to g'(x) = -6/(x-5)^2.

Step 4: Find the Value of the Derivative at the Given Point

The exercise does not specify a point at which to evaluate the derivative. Typically, this information is provided. If you had a particular value of x, you would substitute it into the final expression for the derivative.

Key Concepts

Derivative of a Function

Understanding the concept of a derivative is fundamental in calculus. Essentially, the derivative represents how a function's output value changes as its input value changes. In simpler terms, it is the instantaneous rate of change or slope of the function at any given point. For example, if you graph the function and look at a point on the graph, the derivative tells you how steep the curve is at that point.

For a function defined by an equation, say, \( f(x) = 2x^2 \), the derivative, denoted as \( f'(x) \) or \( \frac{df}{dx} \), can reflect the rate of change at any point \( x \). It is the 'engine' behind differential calculus and is widely used in various fields including physics, engineering, economics, and even medicine.

Applying Differentiation Rules

In calculus, there are specific rules that make finding the derivative of a function more manageable. These rules, including the Power Rule, Product Rule, Chain Rule, and the Quotient Rule, apply to different types of functions. In the case of quotient functions where one function is divided by another, the Quotient Rule is particularly useful. Like a recipe, the Quotient Rule guides us through a step-by-step process. It says that the derivative of a quotient \( \frac{u(x)}{v(x)} \) is \( \frac{u'(x) \cdot v(x) - u(x) \cdot v'(x)}{(v(x))^2} \).

Knowing when and how to apply these rules is a skill developed through practice and understanding the types of functions and their derivatives. Remembering the Quotient Rule, in particular, is essential because it helps in dealing with complex fractions, ensuring that the differentiation process remains systematic and less prone to error.

Calculus in Algebra

Calculus and algebra are distinct fields of mathematics, yet they are deeply intertwined. Calculus often relies on algebraic manipulation to solve problems. When you're differentiating a function, such as \( g(x)=\frac{2x+1}{x-5} \), you start by recognizing the algebraic form of the function—here, it's a ratio of polynomials. Then, you use differentiation rules that stem from algebraic principles to find the derivative.

Within calculus, algebraic operations such as factoring, expanding, and simplifying expressions are pivotal. They enable us to rewrite functions in forms that are more conducive to applying differentiation rules, and they also help us to interpret the resulting derivatives correctly. In the context of our example, we see calculus working hand-in-hand with algebra to simplify the expression after applying the Quotient Rule, leading to the final derivative of \( g'(x)=\frac{-6}{(x-5)^2} \).