Problem 85 - 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 $$ \text { If } f(x)=\sqrt{3 x}, \text { find } \frac{d}{d x}[(f \circ f \circ f)(x)] $$ Do you see a shortcut?

Step-by-Step Solution

Answer

The derivative is \( \frac{3^{5/8}}{4}x^{-3/4} \). A shortcut is recognizing the pattern in iterated compositions form a power function, simplifying derivative calculations.

Step 1: Understanding the Iterated Function

Given the function \( f(x) = \sqrt{3x} \), we need to compute the iterated function \((f \circ f \circ f)(x)\). This means applying \( f \) three times: \( f(f(f(x))) \). Start by applying \( f \) once: \( f(x) = \sqrt{3x} \).

Step 2: First Composition Step

Perform the first composition: \( f(f(x)) = f(\sqrt{3x}) = \sqrt{3(\sqrt{3x})}\). Simplify: \( f(f(x)) = \sqrt{3^{3/2}x^{1/2}} = (3^{1/4})\sqrt{x} \).

Step 3: Second Composition Step

Perform the next composition: \( f(f(f(x))) = f((3^{1/4})\sqrt{x}) \). This results in \( \sqrt{3((3^{1/4})\sqrt{x})} = \sqrt{3^{5/4}x^{1/2}} = (3^{5/8})x^{1/4} \).

Step 4: Applying the Chain Rule

To find the derivative \( \frac{d}{dx}[(f \circ f \circ f)(x)] \) where \( (f \circ f \circ f)(x) = (3^{5/8})x^{1/4} \): Use the derivative formula for power functions, \( \frac{d}{dx} x^n = nx^{n-1} \). For \( x^{1/4} \), the derivative is \( \frac{1}{4}x^{-3/4} \). Therefore, \( \frac{d}{dx}[(3^{5/8})x^{1/4}] = (3^{5/8}) \cdot \frac{1}{4}x^{-3/4} \).

Step 5: Final Calculation of the Derivative

Evaluate the final expression for the derivative: \( \frac{3^{5/8}}{4} x^{-3/4} = \frac{3^{5/8}}{4\sqrt[4]{x^3}} \). Therefore, the final derivative is: \( \frac{3^{5/8}}{4}x^{-3/4} \).

Step 6: Identifying a Shortcut

Notice that each time \( f \) is applied, the exponent and coefficient are modified consistently. By observing the pattern, one can derive similar compositions directly without recalculating the composition, recognizing that the function form (power function in a nested pattern) simplifies the derivative calculation.

Key Concepts

Iterated Functions

Iterated functions involve applying a function repeatedly to its own output. In this concept, the function's result becomes the input for the next operation. This is akin to putting a result back into a machine to see what comes out next. For an iterated function, if we start with a function \( f(x) \), then \((f \circ f)(x)\) means \( f(f(x)) \), or applying \( f \) to \( x \) and then applying \( f \) again to that result. We can also have higher iterations like \((f \circ f \circ f)(x)\), which results in \( f(f(f(x))) \). Iterated functions appear in many real-world applications, from computing compound interest in finance to simulating weather patterns in meteorology. Each application of \( f \) often modifies the function in a predictable manner, providing a foundation for deeper mathematics and complex systems.

Composition of Functions

The composition of functions is a fundamental concept where two functions are combined in such a way that the output of one function becomes the input of another. If you have two functions, \( f(x) \) and \( g(x) \), the composition \((f \circ g)(x)\) means \( f(g(x)) \). We're essentially feeding the output of \( g(x) \) right into \( f(x) \). This process is similar to connecting two pipelines, where the output of the first is routed directly into the intake of the second. In mathematics, this operation is crucial for constructing new functions from existing ones, often leading to more complex and interesting results. Understanding this can be incredibly powerful not only in solving equations but also in modeling sophisticated real-life scenarios.

Chain Rule

The chain rule is vital in calculus for computing derivatives of composite functions. It allows us to find the derivative of a function that is composed of other functions. Imagine nesting one function inside another, much like a matryoshka doll, and wanting to find how sensitively the innermost doll (function) reacts to changes from the outside. For two functions, \( f(u) \) and \( u = g(x) \), the chain rule states that the derivative \( \frac{d}{dx}[f(g(x))] \) is \( f'(g(x)) \cdot g'(x) \). This means you find the derivative of \( f \) with respect to \( u \) and multiply it by the derivative of \( u \) with respect to \( x \). The chain rule is instrumental when handling iterated or composite functions, as it simplifies the differentiation process, providing a much clearer path to the solution.

Derivative Calculation

Calculating derivatives is a primary tool in calculus used to determine the rate of change of a function. When you have an expression like \((f \circ f \circ f)(x)\), accurate derivative calculation helps unravel how this complex function changes as \( x \) changes. For power functions, the derivative formula \( \frac{d}{dx} x^n = nx^{n-1} \) is particularly useful. In our exercise, where \( (f \circ f \circ f)(x) = (3^{5/8})x^{1/4} \), the derivative \( \frac{d}{dx}[(3^{5/8})x^{1/4}] = (3^{5/8}) \cdot \frac{1}{4}x^{-3/4} \) shows us the rate of change. Simplifying gives us \( \frac{3^{5/8}}{4\sqrt[4]{x^3}} \). Being able to calculate derivatives effectively allows us to analyze how function values shift concerning x, providing critical insights into their behavior.