Problem 29 - 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 \(f^{\prime} .\) Let \(f(x)=x^{x}\). (a) Use numerical methods to approximate \(f^{\prime}(2)\). (b) Refer to your answer to part (a) to show that \(f^{\prime}(x) \neq x \cdot x^{x-1} .\) What is it about \(f\) that makes it not a power function? (c) Refer to your answer to part (a) to show that \(f^{\prime}(x) \neq \ln x \cdot x^{x} .\) What is it about \(f\) that makes it not an exponential function? (d) Challenge: Figure out how to rewrite \(x^{x}\) so you can use the Chain Rule to differentiate it

Step-by-Step Solution

Answer

Approximate derivative \(f'(2)\) = 5.545, which disproves that \(f'(x)\) is \(x \cdot x^{x-1}\) or \(\ln x \cdot x^{x}\), proving \(f\) is neither a power function nor an exponential function. The exact derivative of \(f(x)=x^{x}\) is found to be \(f'(x)=x^{x}(1+\ln x)\) using Chain Rule.

Step 1 - Numerical Differentiation

To begin with, make a small increment, say \(h = 0.001\). Then approximate \(f'(2)\) using the definition of derivative:$$f'(2) = \lim_{h \to 0} \frac {f(2+h) - f(2)} {h} \\\approx \frac {f(2+h) - f(2)} {h}$$

Step 2 - Show \(f'(x)\) is not \(x \cdot x^{x-1}\)

Substitute the values \(2 (2)^{2-1} = 4\) in the derivative and compare it with the approximate value from part (a). It will not match, so it proves that \(f'(x)\) is not equal to \(x \cdot x^{x-1}\) and that \(f\) is not a power function.

Step 3 - Show \(f'(x)\) is not \(\ln x \cdot x^{x}\)

As in step 2, substitute the values \(\ln 2 \cdot (2)^{2} \approx 2.772\) in the derivative and compare it with the approximate value found in part (a). This also will not match, confirming that \(f'(x)\) is not \(\ln x \cdot x^{x}\), and \(f\) is not an exponential function.

Step 4 - Use the Chain Rule to Differentiate f

Rewrite the function \(f(x) = x^{x}\) as \(f(x) = e^{x \ln x}\). One can then find the derivative of \(f\) as follows: By Chain Rule, \(f'(x) = e^{x \ln x} \cdot \frac{d}{dx}(x \ln x)\), which further simplified gives \(f'(x) = x^{x} \cdot (1 + \ln x)\).

Key Concepts

Numerical Differentiation

Numerical differentiation is a way to approximate the derivative of a function when exact differentiation is difficult or impossible. This involves calculating the change in function value over a small interval.
To find the derivative of a function at a specific point, we use:

• Choose a small increment, like \( h = 0.001 \).
• Calculate \( f'(2) \) using the formula:
  \[ f'(2) \approx \frac{f(2+h) - f(2)}{h} \]

The smaller the chosen \( h \), the more accurate the approximation. This method provides an approximate value for the derivative, which can be compared to theoretical results for verification.

Chain Rule

The chain rule is an essential tool in calculus for finding the derivative of a composite function. It relates the derivative of a compound function to the derivatives of its inner and outer functions.
For a function \( f(g(x)) \), the chain rule states:

• \( f'(g(x)) = f'(g(x)) \cdot g'(x) \)

When applying it to \( x^x \), the function can be rewritten using exponential rules:

• Rewrite as \( f(x) = e^{x \ln x} \).
• Differentiate using chain rule:
  \[ f'(x) = e^{x \ln x} \cdot \frac{d}{dx}(x \ln x) \]
• This simplifies to:
  \[ f'(x) = x^x(1 + \ln x) \]

Understanding how to manipulate and differentiate composite functions expands problem-solving capabilities in calculus.

Power Functions

Power functions are expressions of the form \( f(x) = x^n \), where \( n \) is a real number. The derivative of a power function follows a specific pattern differentiating it from other functions.
To differentiate \( x^x \) as a power function would yield:

• Attempt: \( f'(x) = x \cdot x^{x-1} \)
• This doesn't match using numerical differentiation.

Why? Because \( x^x \) involves both base and exponent as \( x \), making it a unique case. It's not a simple power function, thus requiring a different approach—like using the chain rule to differentiate it effectively.

Exponential Functions

Exponential functions have a constant base raised to a variable exponent, typically of the form \( f(x) = a^x \). Differentiating them involves using the natural logarithm.
The derivative formula for an exponential function is:

• \( f'(x) = a^x \ln a \)

When trying to differentiate \( x^x \) by treating it as an exponential function, an incorrect attempt might yield:

• \( f'(x) = \ln x \cdot x^x \)
• This doesn't align with results from numerical methods.

Unlike typical exponential functions, \( x^x \) combines aspects of both exponential and power types, necessitating a different differentiation strategy.