Problem 87 - Brief Applied Calculus

Question

PRODUCT RULE FOR THREE FUNCTIONS Show that if \(f, g\), and \(h\) are differentiable functions of \(x\), then $$ \frac{d}{d x}(f \cdot g \cdot h)=f^{\prime} \cdot g \cdot h+f \cdot g^{\prime} \cdot h+f \cdot g \cdot h^{\prime} $$

Step-by-Step Solution

Answer

Differentiate using product rule: \( (f'gh + fg'h + fgh') \).

Step 1: Understand the Product Rule for Two Functions

Before we extend to three functions, recall the product rule for two functions. If you have two differentiable functions, say \( u(x) \) and \( v(x) \), then the derivative of their product is given by: \[ \frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]

Step 2: Apply Product Rule to First Pair of Functions

Consider the functions \( f(x) \), \( g(x) \), and \( h(x) \). Let's first apply the product rule to the product of \( f(x) \cdot g(x) \). Let \( u = f(x) \cdot g(x) \) and \( v = h(x) \), then the derivative of \( u \cdot v \) becomes: \[ \frac{d}{dx}[f \cdot g \cdot h] = \frac{d}{dx}[u \cdot v] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]

Step 3: Differentiate the First Pair of Functions

For the expression \( u = f \cdot g \), apply the product rule again: \[ u'(x) = \frac{d}{dx}[f \cdot g] = f'(x) \cdot g(x) + f(x) \cdot g'(x) \]

Step 4: Substitute Back into the Expression

Substitute the derivative \( u'(x) \) we found into the expression for the derivative of the product \( u \cdot v \):\[ \frac{d}{dx}[f \cdot g \cdot h] = (f'(x) \cdot g(x) + f(x) \cdot g'(x)) \cdot h(x) + f(x) \cdot g(x) \cdot h'(x) \]

Step 5: Simplify

Expand and simplify the expression to achieve the final result:\[ \frac{d}{dx}(f \cdot g \cdot h) = f'(x) \cdot g(x) \cdot h(x) + f(x) \cdot g'(x) \cdot h(x) + f(x) \cdot g(x) \cdot h'(x) \]

Key Concepts

Differentiable Functions

In calculus, a function is said to be differentiable at a certain point if it has a derivative at that point. Determining if a function is differentiable is essential when working with the product rule since it applies to differentiable functions. For a function to be differentiable, it must be continuous. However, continuity alone does not guarantee differentiability. It requires smoothness as well, meaning no sharp corners or cusps at the point in question.
For three functions, such as in the exercise with functions \(f, g,\) and \(h\), each function must be differentiable in order to apply the product rule. Differentiability ensures we can calculate the derivative of each function individually before combining them using calculus rules.

When dealing with multiple functions, remember:

• Each function should be differentiable over the range of interest.
• Ensure that you check for continuity and smooth transitions without abrupt changes.
• Checking differentiability helps prevent errors in derivative calculations later on.

Derivatives

Derivatives are a core concept in calculus, representing the rate at which a function changes. Essentially, the derivative tells you how a function behaves as its input changes, providing insights into the slope or steepness of the curve at any given point.
In the context of the exercise with the product rule, we are concerned with derivatives of products of functions. Calculating derivatives of products involves applying specific rules, notably the product rule. The product rule simplifies the process, allowing us to manage the complexity of multiple function interactions in a systematic way.

To calculate derivatives effectively, consider:

• The definition of a derivative as a limit, showing how it responds to infinitesimal changes in input.
• The importance of understanding which derivative rules apply, such as the product rule for products of functions.
• Using basic derivative rules like power rule, sum rule, and chain rule alongside the product rule to find derivatives of more complex expressions.

Understanding derivatives well lays the groundwork for more advanced calculus topics.

Calculus

Calculus is a branch of mathematics focused on the study of change. It is divided into two main disciplines: differential calculus and integral calculus, each dealing with different aspects of change.
Differential calculus, which deals with derivatives, allows us to understand how functions change and is the framework behind derivative rules, like the product rule demonstrated in the exercise. The product rule is part of a broader set of rules that make calculus possible, enabling us to find slopes of tangents, solve problems related to motion, and optimize functions in real-world applications.

Key calculus concepts include:

• Limit processes, which form the foundation for defining derivatives and integrals.
• Application of rules like the product rule, which handle the differentiation of products of functions.
• Understanding the fundamental theorem of calculus, connecting derivatives and integrals.

By mastering these concepts, calculus becomes a powerful tool, offering solutions to complex problems in physics, engineering, and economics, among many other fields.