Problem 83 - 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 $$ \text { If } f(x)=x+\sqrt{x}, \text { find } \frac{d}{d x}[(f \circ f)(x)] $$

Step-by-Step Solution

Answer

The derivative \(\frac{d}{dx}[(f \circ f)(x)]\) is \(\left(1 + \frac{1}{2\sqrt{x+\sqrt{x}}}\right)\left(1 + \frac{1}{2\sqrt{x}}\right)\)."

Step 1: Find Inner Function

Identify the innermost part of the function. Given \(f(x) = x + \sqrt{x}\), the iterated function is \(f(f(x))\), which means replacing \(x\) in \(f(x)\) with \(f(x)\) itself. Compute \(f(f(x)) = f(x + \sqrt{x}) = (x + \sqrt{x}) + \sqrt{x + \sqrt{x}}\).

Step 2: Simplify Expression

Simplify the expression for \(f(f(x))\). Thus we have, \(f(f(x)) = x + \sqrt{x} + \sqrt{x + \sqrt{x}}\).

Step 3: Apply the Chain Rule

Differentiate \(f(f(x))\) using the Chain Rule. The Chain Rule states \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\). Here, \(g(x) = f(x) = x + \sqrt{x}\) and \(f(x) = x + \sqrt{x}\).

Step 4: Find Derivatives

Differentiate \(f(x) = x + \sqrt{x}\). The derivative is \(f'(x) = 1 + \frac{1}{2\sqrt{x}}\). Set \(g(x) = x + \sqrt{x}\), so that \(g'(x) = 1 + \frac{1}{2\sqrt{x}}\).

Step 5: Substitute into Chain Rule

Using the results from Step 4 in the Chain Rule formula: \(\frac{d}{dx}[(f \circ f)(x)] = f'(g(x)) \cdot g'(x)\). Substitute \(g(x) = x + \sqrt{x}\) into \(f'(x)\), getting \(\left(1 + \frac{1}{2\sqrt{x+\sqrt{x}}}\right)\left(1 + \frac{1}{2\sqrt{x}}\right)\).

Key Concepts

Iterated Functions

When we iterate a function, we compose it with itself multiple times. For example, if we have a function \( f(x) = x + \sqrt{x} \), then the iterated function \((f \circ f)(x)\) becomes \(f(f(x))\). This involves replacing the input of the function \(f(x)\) with the function itself. This operation can be extended to however many iterations needed, such as \((f \circ f \circ f)(x) = f(f(f(x)))\).

Iterated functions prove useful across various domains:

• **Finance**: Compound interest calculations often involve iterating growth formulas.
• **Sciences**: Weather predictions and other scientific models frequently use iterated functions for predictive modeling.

Understanding iterated functions is crucial as they form the basis for more complex relationships explored in calculus and beyond.

Chain Rule

The Chain Rule is a fundamental concept in calculus used for finding the derivative of composite functions. It allows us to differentiate functions that are formed by the composition of two or more functions. Think of it as a tool that helps "unravel" layers of functions to find their rates of change.

This rule is expressed as: \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \] In simple terms, you first differentiate the outer function \(f\), evaluate it at the inner function \(g(x)\), and then multiply by the derivative of the inner function \(g(x)\). This helps you manage situations where direct differentiation isn't straightforward.

• **Common Use**: Often applied in problems involving physical systems where different rates are interconnected.
• **Example**: Differentiating \((f \circ f)(x)\) applies the Chain Rule to find how small changes in \(x\) affect our iterated function.

Without the Chain Rule, you would find it quite challenging to handle differentiated forms of composed functions.

Derivative of Composition

Determining the derivative of a composed function, also known as the derivative of composition, involves using the Chain Rule discussed earlier. The process essentially breaks down a complex function into simpler parts that are easier to analyze.
For example, given the function \( f(x) = x + \sqrt{x} \), its derivative is determined by: \[ f'(x) = 1 + \frac{1}{2\sqrt{x}} \] Let's explore this further with the composition \( (f \circ f)(x) \):

• First, determine \(g(x) = f(x) = x + \sqrt{x} \). This is the "inner" function in your chain.
• The composition \(f(f(x))\) can be broken as \(f(g(x))\).

When we apply the Chain Rule, we find the overall derivative as: \[ \frac{d}{dx}[(f \circ f)(x)] = f'(g(x)) \cdot g'(x) \] Evaluating each function's derivative and substituting them accordingly gives us a complete picture of the rate of change for such compositions. Understanding this derivative method becomes invaluable as you progress into more challenging calculus topics.