Problem 82 - Calculus and its applications

Question

If \(f(x)\) is a function, then \((f \circ f)(x)=f(f(x))\) is the composition of \(f\) with itself. This is called an iterated function, and the composition can be repeated many times. For example, \((f \circ f \circ f)(x)=f(f(f(x))) .\) Iterated functions are very useful in many areas, including finance (compound interest is a simple case) and the sciences (in weather forecasting, for example). For each function, use the Chain Rule to find the derivative $$ \text { If } f(x)=x^{2}+1, \text { find } \frac{d}{d x}[(f \circ f)(x)] $$

Step-by-Step Solution

Answer

The derivative is \(4x^3 + 4x\).

Step 1: Understand the Problem

The problem asks for the derivative of the iterated function \((f \circ f)(x)\), which is given by \(f(f(x))\). The function \(f(x)\) is defined as \(x^2 + 1\).

Step 2: Express the Iterated Function

The iterated function is \((f \circ f)(x) = f(f(x))\). Since \(f(x) = x^2 + 1\), then \(f(f(x)) = f(x^2 + 1)\). Substitute \(x^2 + 1\) into \(f\) to get \(f(x^2 + 1) = (x^2 + 1)^2 + 1\).

Step 3: Apply the Chain Rule

To find \( \frac{d}{dx}[(f \circ f)(x)]\), apply the Chain Rule. Let \(u = f(x) = x^2 + 1\) and therefore \(y = f(u) = u^2 + 1\). The Chain Rule states that \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

Step 4: Compute Derivatives

First, find \( \frac{du}{dx} \), which is the derivative of \(x^2 + 1\) with respect to \(x\), giving \(2x\). Next, find \( \frac{dy}{du} \), which is the derivative of \(u^2 + 1\) with respect to \(u\), providing \(2u\).

Step 5: Substitute and Multiply

Substitute \(u = x^2 + 1\) into \( \frac{dy}{du} = 2u\), making \( \frac{dy}{du} = 2(x^2 + 1)\). Now multiply \( \frac{dy}{du} = 2(x^2 + 1)\) by \( \frac{du}{dx} = 2x\), resulting in \( \frac{dy}{dx} = 2(x^2 + 1) \cdot 2x = 4x(x^2 + 1)\).

Step 6: Simplify the Expression

Expand the expression \(4x(x^2 + 1)\) to get \(4x^3 + 4x\). Thus, the derivative of \((f \circ f)(x)\) is \(4x^3 + 4x\).

Key Concepts

Chain Rule

The Chain Rule is a fundamental technique in calculus for differentiating composite functions. When dealing with a composite function, you are essentially looking at a function within another function. The Chain Rule helps to unravel these layers by focusing on how the inner function affects the outer function.
To apply the Chain Rule, first identify the inside and the outside functions. Suppose you have a function formed as \(y = f(g(x))\). The rule states:

• Find the derivative of the outer function while treating the inside function as a variable. This means taking \(\frac{dy}{du}\) where \(u = g(x)\).
• Next, find the derivative of the inner function \(g(x)\) with respect to \(x\), which is \(\frac{du}{dx}\).
• The final step is to multiply these two derivatives together: \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

By systematically breaking down the composite function, the Chain Rule allows us to calculate the derivative accurately. It is particularly handy when dealing with iterated functions, where a function repeatedly applies to itself.

Iterated Functions

Iterated functions involve applying a function to its own output repeatedly. This creates a series of compositions where each result feeds into the next iteration. In mathematical terms, if you start with a function \(f(x)\), the iterated functions will be \((f \circ f)(x) = f(f(x))\), \((f \circ f \circ f)(x)\), and so on.
This recursive application makes iterated functions unique tools in various fields like finance and physics. For example, in finance, compound interest calculations rely on iterating interest over time. In physics, iterated functions appear in models like weather predictions where previous states affect future outcomes. Understanding how to approach and calculate derived iterated functions is essential when utilizing them in practical applications.

Derivative

The derivative is a measure of how a function changes as its input changes. It gives the rate at which the function's value is changing at any given point. In essence, the derivative represents the slope of the function's graph.
To calculate the derivative, you take a small change in \(x\), called \(dx\), and look at how the function value changes, denoted \(dy\). Then, \(\frac{dy}{dx}\) provides the derivative of the function with respect to \(x\). For polynomials, the derivative is obtained by multiplying the exponent by the coefficient and subtracting one from the exponent, such as \(\frac{d}{dx} x^n = nx^{n-1}\).
This serves as the foundation of calculus, providing insights into the behavior and dynamics of functions, which is particularly beneficial in optimization and modeling.

Function Composition

Function composition is the process of applying one function to the result of another function. This is expressed as \((f \circ g)(x) = f(g(x))\). The order of application is crucial since \(f(g(x))\) may yield a different result than \(g(f(x))\).
Composing functions helps to build more complex functions from simpler ones, enabling us to model intricate real-world scenarios. By chaining operations, such as adding complexity to a physical model or refining calculations within computer programs, function composition is vital.
When dealing with derivatives of composed functions, the Chain Rule (as discussed) becomes the go-to tool, ensuring that we evaluate the relationship between the nested functions correctly, allowing us to find the rate of change or optimization of the composed structure.