Problem 43 - Brief Applied Calculus

Question

Find the derivative of each function by using the Quotient Rule. Simplify your answers. $$ f(x)=\frac{x^{4}+x^{2}+1}{x^{2}+1} $$

Step-by-Step Solution

Answer

The derivative of the function is \( f'(x) = \frac{2x^5 + 4x^3}{(x^2 + 1)^2} \).

Step 1: Identify the functions

Identify the functions for which we'll use the Quotient Rule. We have the numerator as \( u(x) = x^4 + x^2 + 1 \) and the denominator as \( v(x) = x^2 + 1 \).

Step 2: Write down the Quotient Rule formula

Recall the Quotient Rule, which is given by: \[ \left( \frac{u(x)}{v(x)} \right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]. We will apply this formula to find the derivative of the given function.

Step 3: Differentiate the numerator and the denominator

Differentiate the numerator function: \( u(x) = x^4 + x^2 + 1 \), which gives \( u'(x) = 4x^3 + 2x \). Differentiate the denominator function: \( v(x) = x^2 + 1 \), which gives \( v'(x) = 2x \).

Step 4: Substitute into the Quotient Rule formula

Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the Quotient Rule: \[ f'(x) = \frac{(4x^3 + 2x)(x^2 + 1) - (x^4 + x^2 + 1)(2x)}{(x^2 + 1)^2} \].

Step 5: Expand and simplify the numerator

Expand the terms in the numerator: - \((4x^3 + 2x)(x^2 + 1) = 4x^5 + 4x^3 + 2x^3 + 2x = 4x^5 + 6x^3 + 2x\)- \((x^4 + x^2 + 1)(2x) = 2x^5 + 2x^3 + 2x\)Subtract to get: \(4x^5 + 6x^3 + 2x - (2x^5 + 2x^3 + 2x) = 2x^5 + 4x^3\).

Step 6: Write the final derivative

The simplified derivative is \[ f'(x) = \frac{2x^5 + 4x^3}{(x^2 + 1)^2} \], which is the result of using and simplifying the Quotient Rule on the given function.

Key Concepts

Derivatives

Derivatives are a fundamental concept of calculus and describe how a function changes as its input changes. At the core, the derivative measures the rate of change of a quantity. It's like watching a car's speed on a dashboard to see how fast it's going at any point.
In mathematical terms, when you have a function, the derivative tells you the slope of the tangent line to the graph of the function at a particular point. This is extremely useful in predicting how a function behaves under various conditions.

• It helps us find maximum and minimum points of functions.
• It aids in determining points of inflection where curves change concavity.
• Applications include physics for velocity and acceleration, economics for cost and revenue rates, and many other fields.

Understanding derivatives is crucial for anyone looking to delve deeper into calculus and applied mathematics.

Function Differentiation

Function differentiation is the process of finding the derivative of a function. This can be done using several rules and techniques, including the Quotient Rule, which is particularly useful when dealing with rational functions, or functions that can be expressed as a fraction of two polynomials.
When differentiating, we apply specific rules to obtain the derivative. For instance, the Power Rule simplifies finding derivatives of functions like polynomials. The Quotient Rule, used in this exercise, is essential when the function is a ratio, such as \[ rac{u(x)}{v(x)} \]. With these tools, you can tackle various types of functions:

• Quotient Rule for functions divided by another.
• Chain Rule for nested or composed functions.
• Product Rule when dealing with products of two functions.

Practicing these rules helps in quickly differentiating and simplifying even the most complex expressions.

Calculus

Calculus is the branch of mathematics dealing with continuous change. It is divided into two main parts: differential calculus and integral calculus. While derivative calculus (our main focus here) deals with finding the derivative or the rate of change, integral calculus deals with finding the integral or the cumulative sum of quantities.
In our specific task of using the Quotient Rule, we are diving into differential calculus. This allows us to understand how a function like \( f(x) = \frac{x^4 + x^2 + 1}{x^2 + 1} \) behaves. By applying the rules of calculus, one can learn to

• Calculate instantaneous rates of change.
• Understand the behavior of curves and surfaces.
• Apply these concepts to fields ranging from science to engineering.

Calculus not only provides the tools to solve mathematical problems but also offers insight into the natural and technological world. Embracing its concepts opens up a myriad of possibilities for exploration and discovery.