Problem 32 - Brief Applied Calculus

Question

Find the derivative of each function by using the Quotient Rule. Simplify your answers. $$ f(x)=\frac{x^{5}-1}{x^{2}} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 3x^2 + \frac{2}{x^3} \).

Step 1: Identify the Functions

In the function \( f(x) = \frac{x^5 - 1}{x^2} \), identify the numerator and the denominator. Here, the numerator is \( u(x) = x^5 - 1 \) and the denominator is \( v(x) = x^2 \).

Step 2: Find the Derivatives

Calculate the derivative of the numerator \( u'(x) \) and the derivative of the denominator \( v'(x) \). For \( u(x) = x^5 - 1 \), \( u'(x) = 5x^4 \). For \( v(x) = x^2 \), \( v'(x) = 2x \).

Step 3: Apply the Quotient Rule Formula

The Quotient Rule states that the derivative of a function \( f(x) = \frac{u(x)}{v(x)} \) is \( f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \). Substitute \( u(x) \), \( v(x) \), \( u'(x) \), and \( v'(x) \) into this formula.

Step 4: Substitute and Simplify

Substitute into the Quotient Rule: \( f'(x) = \frac{x^2 \, (5x^4) - (x^5 - 1) \, (2x)}{(x^2)^2} \). This becomes \( f'(x) = \frac{5x^6 - 2x(x^5) + 2x}{x^4} \).

Step 5: Simplify the Expression Further

Simplify the expression: \( 5x^6 - 2x^6 + 2x = 3x^6 + 2x \). So the numerator is \( 3x^6 + 2x \) and the denominator is \( x^4 \).

Step 6: Final Simplified Derivative

Divide each term in the numerator by \( x^4 \): \( f'(x) = \frac{3x^6}{x^4} + \frac{2x}{x^4} = 3x^2 + \frac{2}{x^3} \). This is the simplified derivative of the function.

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus. They represent how a function changes or "rates of change." When we talk about derivatives, we often refer to the slope of the tangent line at any given point on the graph of the function.

For a simple function like a straight line, the derivative is constant since the line's slope remains the same throughout. However, when a function is more complex, like a polynomial or a quotient of functions, finding the derivative requires special rules, such as the Quotient Rule used in this exercise.

The derivative gives you practical information, such as:

• The speed of an object as it travels over time (in physics).
• The rate at which revenue changes as you adjust prices (in economics).

Learning how to compute derivatives of different functions helps you understand the real-world phenomena that these functions model.

Simplifying Expressions

The process of simplifying expressions in calculus usually involves combining like terms or reducing fractions. Once you calculate the derivative using a rule such as the Quotient Rule, you often end up with a complex expression.

Here are some common techniques for simplifying expressions:

• Combine like terms: Look for terms with the same variables raised to the same power and add or subtract them.
• Cancel terms: Divide both the numerator and denominator of a fraction by their greatest common divisor.
• Factor terms: Factor out common variables or numbers to simplify expressions.

Simplification is crucial not only for making your results easier to interpret but also for spotting errors and verifying the correctness of your initial differentiation.

Calculus Step-by-Step with the Quotient Rule

Solving calculus problems step-by-step ensures you fully understand each part of the process. The Quotient Rule is especially useful for functions that are defined as one polynomial divided by another.

Here’s how you solve such problems using the Quotient Rule:

• Identify the numerator and denominator. In the problem given, the numerator is \( x^5 - 1\) and the denominator is \( x^2 \).
• Find the derivative of both the numerator and the denominator. The derivative of the numerator \( x^5 - 1 \) is \( 5x^4 \), and the derivative of the denominator \( x^2 \) is \( 2x \).
• Apply the Quotient Rule formula: \( f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \).
• Substitute the functions and their derivatives into the formula.
• Finally, simplify the resulting expression to find the most reduced form of the derivative.

Working through these steps not only helps you solve the problem at hand but also builds your confidence in handling more complex calculus problems in the future.