Problem 44 - College Algebra and Calculus: An Applied Approach

Question

find \(f^{\prime}(x)\). $$ f(x)=\frac{2 x^{2}-3 x+1}{x} $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x)=\frac{2 x^{2}-3 x+1}{x}\) is \(f'(x)= \frac{2x^{2} - 1}{x^{2}}\).

Step 1: Define the Functions

First, let's specify the functions. We have two functions divided, let \(u(x) = 2x^{2} - 3x +1\) be the numerator and let \(v(x) = x\) be the denominator.

Step 2: Compute Derivatives of u and v

Next, calculate the derivatives of \(u(x)\) and \(v(x)\). The derivative of \(u(x)\) can be found by applying the power rule for derivatives, which yields \(u'(x) = 4x - 3\). For \(v(x)\), since it's simply x, its derivative, \(v'(x)\), is 1.

Step 3: Apply the Quotient Rule for Derivatives

Now, we use the quotient rule as follows: if we have a function in the form \(\frac{u(x)}{v(x)}\), its derivative, \(f'(x)\), is given by \(\frac{u'(x)v(x) - u(x)v'(x)}{v(x)^{2}}\). Substituting our functions and their derivatives into this formula, we get \(f'(x) = \frac{(4x - 3)x - (2x^{2} - 3x +1) \cdot 1}{x^{2}}\).

Step 4: Simplify the Expression

Simplify the above expression to obtain \(f'(x) = \frac{4x^{2} - 3x - 2x^{2} + 3x -1}{x^{2}} = \frac{2x^{2} - 1}{x^{2}}\).

Key Concepts

Calculus

Calculus is a fundamental branch of mathematics focused on the study of changes. Central to calculus are the concepts of limits, derivatives, integrals, and the Fundamental Theorem of Calculus. Calculus allows us to understand how quantities change and to calculate rates of change and areas under curves. It's an essential tool in sciences and engineering for analyzing dynamic systems and physical phenomena.

Those engaging with calculus will often encounter different types of functions and will need to determine their derivatives. The derivative is a measure of how a function's output value changes concerning changes in the input value. Knowing how to find a derivative is crucial, and there are specific rules, like the quotient rule for derivatives and the power rule, that help us calculate the derivative of complex functions effectively.

Derivative of a Quotient

The derivative of a quotient refers to finding the derivative when our function is a division of two other functions. There is a specific process for this situation called the quotient rule. In mathematical terms, if you have a function that can be expressed as \(f(x) = \frac{u(x)}{v(x)}\), then the derivative of \(f\) is found using the quotient rule formula \(f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^{2}}\).

To apply this rule effectively, we first need to identify the top part of the division, \(u(x)\), and the bottom part, \(v(x)\), and separately find their derivatives, \(u'(x)\) and \(v'(x)\). Once we have these derivatives, we substitute them into the quotient rule formula to find the derivative of the entire function.

• Identify \(u(x)\) and \(v(x)\).
• Calculate \(u'(x)\) and \(v'(x)\).
• Plug these into the quotient rule formula.

Power Rule for Derivatives

The power rule is a basic principle that simplifies the process of taking the derivative of functions where the variable \(x\) is raised to a power. According to the power rule, if a function is of the form \(f(x) = x^{n}\), where \(n\) is a real number, the derivative of the function, \(f'(x)\), is \(nx^{n-1}\). This rule helps in quickly finding derivatives without resorting to the definition of the derivative as the limit of a ratio of differences.

Using the power rule, we were able to find the derivative of the numerator function \(u'(x)\) in the exercise by applying it to each term of \(u(x) = 2x^{2} - 3x + 1\). This gives \(u'(x) = (2 \cdot 2)x^{2-1} - (3 \cdot 1)x^{1-1} = 4x - 3\). The power rule speeds up the process of finding derivatives and allows us to handle polynomial functions with ease.

Simplifying Expressions

Simplifying expressions is a useful technique in calculus to make equations more manageable and to reveal their basic form. Simplification may involve combining like terms, factoring, expanding expressions, or canceling common factors in fractions. The primary goal is to reduce the expression to its simplest form to facilitate further operations, like taking derivatives or solving equations.

In the context of our exercise, simplifying the expression after applying the quotient rule was the final step. We combined like terms and canceled common factors, which led to the simpler derivative \(f'(x) = \frac{2x^{2} - 1}{x^{2}}\). Simplifying makes it easier to interpret the result and perform any additional required calculations.