Problem 27 - The Calculus with Analytic Geometry

Question

If \(f, g\), and \(h\) are functions and if \(\phi(x)=f(x) \cdot g(x) \cdot h(x)\), prove that if \(f^{\prime}(x), g^{\prime}(x)\), and \(h^{\prime}(x)\) exist, \(\phi^{\prime}(x)=\) \(f(x) \cdot g(x) \cdot h^{\prime}(x)+f(x) \cdot g^{\prime}(x) \cdot h(x)+f^{\prime}(x) \cdot g(x) \cdot h(x)\). (HINT: Apply Theorem 3.5.6 twice.)

Step-by-Step Solution

Answer

The derivative is \(\phi^{\prime}(x) = f(x) \cdot g(x) \cdot h^{\prime}(x) + f(x) \cdot g^{\prime}(x) \cdot h(x) + f^{\prime}(x) \cdot g(x) \cdot h(x)\).

Step 1: Understand the Problem

The task is to prove the derivative of the product of three functions. The given functions are: \(f(x)\), \(g(x)\), and \(h(x)\), and we need to find the derivative of \(\phi(x) = f(x) \cdot g(x) \cdot h(x)\).

Step 2: Application of Theorem 3.5.6

Theorem 3.5.6 states the Product Rule for differentiation: \( (uv)' = u'v + uv' \). To solve this, it will be applied twice due to the involvement of three functions.

Step 3: Apply Product Rule to Two Functions

Consider \(u(x) = f(x) \cdot g(x)\) and \(v(x) = h(x)\). Using the Product Rule on \(\phi(x) = u(x) \cdot v(x)\), we get: \(\phi^{\prime}(x) = u^{\prime}(x) \cdot v(x) + u(x)'v^{\prime}(x)\).

Step 4: Differentiate \(u(x) = f(x) \cdot g(x)\)

Next, we need to find \(u^{\prime}(x)\), which is the derivative of \(f(x) \cdot g(x)\). Apply the Product Rule again: \(u^{\prime}(x) = f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x)\).

Step 5: Substitute and Simplify

Substitute \(u^{\prime}(x)\) back into the expression for \(\phi^{\prime}(x)\). Now: \[\phi^{\prime}(x) = \left( f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x) \right) \cdot h(x) + \left( f(x) \cdot g(x) \right) \cdot h^{\prime}(x)\].

Step 6: Final Simplification

Expand and simplify the expression: \[\phi^{\prime}(x) = f^{\prime}(x) \cdot g(x) \cdot h(x) + f(x) \cdot g^{\prime}(x) \cdot h(x) + f(x) \cdot g(x) \cdot h^{\prime}(x)\].

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that measures how a function changes as its input changes. The derivative of a function gives us the rate at which the function value is changing at any given point. For example, if you have a function \( f(x) \), its derivative is denoted by \( f'(x) \) or \( \frac{d}{dx}f(x) \). In simple terms, if \( f(x) \) represents the position of a car at time \( x \), then \( f'(x) \) would be the speed of the car at time \( x \).

The process of finding a derivative is called differentiation, and it involves applying specific rules (like the Product Rule, Quotient Rule, and Chain Rule) to compute the derivative based on the function's structure. Differentiation has numerous applications across science, engineering, and economics, often helping to optimize processes and understand dynamic systems.

Three Function Product Rule

The Product Rule is an essential tool for differentiating products of functions. When you have three functions \( f(x) \), \( g(x) \), and \( h(x) \), the derivative of their product \( \phi(x) = f(x) \cdot g(x) \cdot h(x) \) requires a careful approach.

When differentiating a product of three functions, we apply the Product Rule more than once. Specifically, we break the problem down step-by-step:

• First, treat the product of two functions (say \( f(x) \text{ and } g(x) \)) as a single function.
• Next, apply the Product Rule to differentiate this new function times the third function (\( h(x) \)).
• This gives us a derivative that considers each function's contribution to the overall rate of change.

The formula for differentiating the product of three functions, \( \phi'(x) = f'(x) \cdot g(x) \cdot h(x) + f(x) \cdot g'(x) \cdot h(x) + f(x) \cdot g(x) \cdot h'(x) \), ensures each function's derivative gets correctly incorporated.

This method maintains order and clarity, making it easier to handle more complex differentiation problems.

Theorem Application

Applying theorems correctly is crucial in mathematics, especially in calculus. When differentiating the product of three functions, we use Theorem 3.5.6, which is the Product Rule for differentiation. Its basic form is: \( (uv)' = u'v + uv' \), where \( u \) and \( v \) are functions of \( x \).

To handle three functions \( f(x) \), \( g(x) \), and \( h(x) \), we:

• Consider two functions at a time, essentially reducing the problem temporarily before expanding it again.
• First, treat \( f(x) \cdot g(x) \) as one function, then differentiate the product of this function with \( h(x) \).
• After finding derivatives, add them up by plugging back into the main formula.

Here, the application of the theorem involves substituting and simplifying stepwise. By understanding each application step, you better grasp the systematic approach to solving complex differentiation problems. This meticulous breakdown will strengthen your foundational skills in calculus and help solve various practical problems.