Problem 28 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 23 through 29, differentiate. In Problems 23 through 25, assume \(f\) is differentiable. Your answers may be in terms of \(f\) and \(f^{\prime} .\) $$ f(x)=\left(x^{3}+e\right)^{\pi} $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = (x^{3} + e)^{\pi}\) is \(f'(x) = \pi(x^{3}+e)^{\pi-1} * 3x^{2}\).

Step 1: Identify the Outer Function

Recognize the structure of the problem as an outer function. Here, the outer function is \(g(u)=u^{\pi}\) where \(u=wx^{3}+e\).

Step 2: Identify the Inner function

The inner function is \(h(x) = x^{3}+e\). This function is nested inside the outer function.

Step 3: Differentiate the Outer Function with respect to u

Now differentiate the outer function \(g(u) = u^{\pi}\). The derivative is \(g'(u)= \pi u^{\pi-1}\).

Step 4: Differentiate the Inner function with respect to x

Then differentiate the inner function \(h(x) = x^{3} + e\), the derivative of which is \(h'(x) = 3x^{2}\). Note that the \(e\) is a constant, and the derivative of a constant is 0.

Step 5: Apply the Chain Rule

The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. Therefore, \(f'(x) = g'(u) * h'(x) = \pi(x^{3}+e)^{\pi-1} * 3x^{2}\).

Key Concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus used for finding the derivative of composite functions. When we have a function composed of another function, the Chain Rule allows us to differentiate them efficiently. The formal definition states that if we have two functions, say \(g(u)\) and \(h(x)\), then the derivative of the composite function \(f(x) = g(h(x))\) is given by

• \(f'(x) = g'(h(x)) \cdot h'(x)\).

In simple terms, the Chain Rule tells us to take the derivative of the outer function (\(g\)) evaluated at the inner function (\(h\)) and multiply it by the derivative of the inner function (\(h\)).
The Chain Rule is essential because it simplifies the process of differentiation when dealing with complex functions built by combining simpler functions. Think of it as peeling an onion: you handle the outer layer first, then work your way inward. This approach is particularly useful in real-life problems where functions tend to be layers of operations built on top of each other.

Composite Functions

Composite functions occur when one function is applied to the result of another function. Practically, this means you have an inner function inside an outer function. For example, given \(f(x) = (x^3 + e)^{\pi}\), we identify two layers:

• The inner function: \(h(x) = x^3 + e\)
• The outer function: \(g(u) = u^{\pi}\)

Here, \(f(x)\) is the composite of \(g\) and \(h\), written as \(f(x) = g(h(x))\).
Recognizing composite functions is crucial as it dictates the approach to differentiate them. Composite functions are commonly seen in calculus problems, which often combine basic algebraic, trigonometric, or exponential functions. Learning to untangle these layers enables better problem-solving skills and helps in applying methods like the Chain Rule effectively.

Derivative Calculation

Calculating derivatives is a core part of calculus that allows us to find the rate at which a function changes. It's essentially finding the slope of the function at any given point. Let's see how it’s done for composite functions using the exercise:Once we identify the composite structure, we differentiate each layer:

• Differentiate the outer function: \(g(u) = u^{\pi}\) becomes \(g'(u) = \pi u^{\pi - 1}\).
• Differentiate the inner function: \(h(x) = x^3 + e\), where the derivative \(h'(x) = 3x^2\).

Applying the Chain Rule, the derivative of the original composite function \(f(x)\) is

• \(f'(x) = g'(h(x)) \cdot h'(x) = \pi (x^3 + e)^{\pi - 1} \cdot 3x^2\).

By breaking down the function into layers and using the Chain Rule, the previously complex-looking expression becomes manageable. Understanding derivative calculation is fundamental to addressing questions around rates of change, optimization, and modeling real-world phenomena in calculus.