Question

Using the Product Rule In Exercises 1-6, use the Product Rule to find the derivative of the function. $$ g(x)=\left(x^{2}+3\right)\left(x^{2}-4 x\right) $$

Short answer

The derivative \(g'(x)\) of the given function \(g(x)\) is \(4x^{3} - 8x^{2} - 12x\).

Step-by-step solution

Understand the Product Rule

The Product Rule is a formula used to find the derivative of a product of two (or more) functions. If two functions u(x) and v(x) are differentiable (i.e. their derivative exists), then the derivative of their product is calculated through the formula: \((u(x) \cdot v(x))' = u'(x) \cdot v(x) + u(x) \cdot v'(x)\). The first thing we need to do is identify the two function that are being multiplied. Here we have \(u(x) = x^{2} + 3\), and \(v(x) = x^{2} - 4x\).

Take the derivative of each function

We apply the rules of derivatives to find the derivatives of u and v. The derivative of \(u(x) = x^{2} + 3\) with respect to x is \(u'(x) = 2x\), and the derivative of \(v(x) = x^{2} - 4x\) with respect to x is \(v'(x) = 2x - 4\).

Apply the Product Rule

We substitute the original functions \(u(x)\) and \(v(x)\) and their derivatives \(u'(x)\) and \(v'(x)\) into the Product Rule equation \((u \cdot v)' = u' \cdot v + u \cdot v'\). This gives us \((u \cdot v)' = 2x \cdot (x^{2} - 4x) + (x^{2} + 3) \cdot (2x - 4)\).

Simplify the expression

We can simplify this expression by distributing the multiplication in each term. This gives: \(2x^{3} - 8x^{2} + 2x^{3} - 12x\). Further simplifying by combining like terms gives us our final answer. The solution for the derivative \(g'(x)\) is \(4x^{3} - 8x^{2} - 12x\).

Key concepts

Derivative

A derivative is a fundamental concept in calculus that represents how a function changes as its input changes. In simpler terms, it tells us the rate at which one quantity changes with respect to another. Imagine driving a car—while the speedometer shows you the speed at that exact moment, the derivative reflects this idea in mathematics. For a given function, the derivative at any point offers a way to estimate how the function behaves at that particular point.

To find the derivative of a function, we can apply various rules, depending on the function's composition. These rules include the Power Rule, Product Rule, Chain Rule, and others. Here, we focus on the Product Rule, which is essential while dealing with products of functions.

Differentiable functions

For the derivative to exist at a point, a function must be differentiable at that point. This means the function has no sharp corners or vertical tangents at that point. Instead, it should be smooth, allowing us to trace a single, unbroken curve that represents the function.

Functions that are polynomial, such as our example functions, tend to be differentiable everywhere because they are smooth and continuous. In the given exercise, the functions in question are polynomials:

• v(x) = x^2 - 4x

These remain differentiable across all real numbers, enabling us to apply the Product Rule confidently.

Simplification of expressions

After applying the Product Rule, we often end up with a complex expression. The next step is simplification. Simplifying expressions involves techniques such as distributing terms, combining like terms, and factoring, which help us express our results in the simplest form possible.

In this exercise, after finding the derivative using the Product Rule, it results in the expression:

• 2x^3 - 8x^2 + 2x^3 - 12x

Next, we simplify by combining like terms.

• The terms 2x^3 and 2x^3 add up to 4x^3
• While -8x^2 stands alone, and -12x is by itself as well.

This gives us a neat expression: 4x^3 - 8x^2 - 12x, which is the simplified expression for the derivative g'(x).

Calculus

Calculus might seem intimidating, but it's actually a fascinating branch of mathematics that allows us to study change. It helps us understand and model real-life phenomena, like the growth of populations or the movements of planets. Calculus is broadly divided into two main branches:

• Differential Calculus - focuses on the concept of the derivative, which we've discussed. It helps us understand rates of change and slopes of curves.
• Integral Calculus - deals with the opposite process of finding areas under curves and accumulations.

In the context of this exercise, differential calculus is crucial. By using rules like the Product Rule, we manage to find derivatives efficiently.

Understanding these rules can simplify complex problems, making calculus not just manageable but also deeply insightful. This exercise connects these theoretical ideas to practical problem-solving by taking theoretical principles, applying them, and then using simplification to find a clear solution.