Question

Suppose that \(f\) is a differentiable function. (a) Find \(\frac{d}{d x} f(f(x))\). (b) Find \(\frac{d}{d x} f(f(f(x)))\). (c) Let \(f^{[n]}\) denote the function defined as follows: \(f^{[1]}=f\) and \(f^{[n]}=f \circ f^{[n-1]}\) for \(n \geq 2 .\) Thus \(f^{[2]}=f \circ f, f^{[3]}=\) \(f \circ f \circ f\), etc. Based on your results from parts (a) and (b), make a conjecture regarding \(\frac{d}{d x} f^{[n]} .\) Prove your conjecture.

Short answer

(a) \( f'(f(x)) \cdot f'(x) \); (b) \( f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) \); (c) \( \prod_{i=1}^{n} f'(f^{[i-1]}(x)) \).

Step-by-step solution

Solve part (a) using the Chain Rule

To solve \( \frac{d}{d x} f(f(x)) \), we apply the chain rule, which states that if you have a composition of functions \( g(h(x)) \), then the derivative is \( g'(h(x)) \cdot h'(x) \).Apply this to \( f(f(x)) \):1. Let \( u = f(x) \).2. The derivative of \( f(u) \) with respect to \( u \) is \( f'(u) \).3. The derivative of \( u = f(x) \) with respect to \( x \) is \( f'(x) \).Thus, \( \frac{d}{d x} f(f(x)) = f'(f(x)) \cdot f'(x) \).

Solve part (b) using repeated application of the Chain Rule

For \( \frac{d}{d x} f(f(f(x))) \), we need to apply the chain rule twice:1. Let \( u = f(f(x)) \).2. The derivative of \( f(u) \) with respect to \( u \) is \( f'(u) \).3. To find \( \frac{d u}{d x} \), note that \( u = f(f(x)) \). We found in part (a) that \( \frac{d u}{d x} = f'(f(x)) \cdot f'(x) \).Thus, \( \frac{d}{d x} f(f(f(x))) = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) \).

Make a conjecture for part (c)

Based on results from parts (a) and (b), we observe a pattern in the derivatives.(a) For \( f(f(x)) \), the derivative is \( f'(f(x)) \cdot f'(x) \).(b) For \( f(f(f(x))) \), the derivative is \( f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) \).Each derivative involves multiplying \( f'(\cdot) \) evaluated at the previous iteration. Therefore, the conjecture for \( \frac{d}{d x} f^{[n]}(x) \) is a product of \( n \) terms: \[ f'(f^{[n-1]}(x)) \cdot f'(f^{[n-2]}(x)) \cdot \ldots \cdot f'(x). \]

Prove the conjecture using mathematical induction

To formally prove this conjecture, use mathematical induction:**Base case**: For \( n = 1 \), \( f^{[1]}(x) = f(x) \). The derivative is \( f'(x) \), which matches our formula as it is simply one term.**Inductive step**: Assume that for some \( k \geq 1 \), the conjecture holds for \( f^{[k]}(x) \), that is, \( \frac{d}{d x} f^{[k]}(x) = f'(f^{[k-1]}(x)) \cdot f'(f^{[k-2]}(x)) \cdot \ldots \cdot f'(x) \).Now prove it for \( n = k + 1 \):1. \( f^{[k+1]}(x) = f(f^{[k]}(x)) \).2. Using the chain rule, \( \frac{d}{d x} f^{[k+1]}(x) = f'(f^{[k]}(x)) \cdot \frac{d}{d x} f^{[k]}(x) \).3. By the inductive hypothesis, \( \frac{d}{d x} f^{[k]}(x) = f'(f^{[k-1]}(x)) \cdot \ldots \cdot f'(x) \).Putting it together:\[ \frac{d}{d x} f^{[k+1]}(x) = f'(f^{[k]}(x)) \cdot f'(f^{[k-1]}(x)) \cdot \ldots \cdot f'(x), \]which completes the inductive step.Thus, the conjecture is proven by induction.

Key concepts

Differentiable Function

A differentiable function is a crucial concept in calculus that uses the idea of derivatives. If a function is differentiable at a point, it means we can calculate its derivative at that point. The derivative tells us the rate at which the function is changing at that specific point.
Think about a differentiable function like a road that is smooth everywhere; you can measure how steep the road is as you go along. Mathematically, this means that the function has a derivative that exists everywhere in its domain where it is differentiable.
The concept of differentiability ensures that the function is continuous. However, the reverse isn't true; a function being continuous doesn't necessarily mean it's differentiable. For example, a function may curve sharply or have a 'corner,' where you can see the function is continuous but not smooth, and hence not differentiable.
Differentiable functions are often used in:

• Optimization problems
• Modeling real-world dynamic systems
• Finding local maxima and minima using techniques like the derivative test

Understanding how a function behaves and changes is vital, and recognizing whether it's differentiable gives us a powerful tool in calculus.

Mathematical Induction

Mathematical induction is a method to prove statements or formulas that are asserted to be true for all natural numbers. Imagine it like a series of dominoes: if you can knock the first one over, and show that knocking one over means the next one will also fall, then all the dominoes will eventually fall. This process involves two main steps:
1. **Base Case:** First, you prove that the statement is true for the initial value, typically 0 or 1. This is like making sure the first domino can indeed be knocked over.
2. **Inductive Step:** Next, you assume the statement is true for some arbitrary natural number, say \( k \), and then prove it's also true for \( k+1 \). This is like ensuring if one domino falls, it causes the next domino to fall.
When both steps are successfully demonstrated, it confirms the statement is true for all natural numbers. Induction is particularly useful for proving formulae and properties involving sequences or recursively defined functions, like in the exercise above where we proved the conjecture regarding the chain of derivatives.

Composition of Functions

The composition of functions is a process where you apply one function to the result of another function. If you think of functions as machines, where you put an input in to receive an output, a composition of functions is like passing your output from the first machine (function) into the next.
Mathematically, if you have two functions, \( f \) and \( g \), the composition is written as \( (f \circ g)(x) = f(g(x)) \). This means you first apply \( g \) to \( x \), then apply \( f \) to the result of \( g(x) \).
Understanding the composition of functions is essential when working with derivatives, especially in applying rules like the Chain Rule. The Chain Rule is specifically about the derivatives of composed functions, telling us how to manage the layering of functions mathematically.
Compositions are seen in various mathematical scenarios:

• Creating complex models in sciences
• Building mathematical algorithms
• In computer science functions, combining outputs of different operations

Comprehending how functions interact and combine to form new functions enriches your calculus toolkit, enabling you to tackle more intricate problems.