Problem 33 - Calculus

Question

Estimate the derivative of \(f(x)=x^{x}\) at \(x=2\).

Step-by-Step Solution

Answer

The derivative of \(f(x) = x^x\) at \(x = 2\) is approximately 8.14.

Step 1: Set Up the Derivative

We want to estimate the derivative of the function \(f(x) = x^x\) at \(x = 2\). First, recall the definition of the derivative: \(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\).

Step 2: Calculate Function Values

Plug into the function \(f(x) = x^x\) to find \(f(2)\) and \(f(2 + h)\). The function value at \(x = 2\) is \(f(2) = 2^2 = 4\). For \(f(2 + h)\), it is \((2+h)^{(2+h)}\).

Step 3: Simplify Expression

Substitute these values into the derivative definition: \[\frac{f(2+h) - f(2)}{h} = \frac{(2+h)^{(2+h)} - 4}{h}.\]This complex expression represents the change in function value around \(x = 2\).

Step 4: Use Logarithmic Differentiation (Optional)

To simplify further, consider using logarithms to differentiate, as \(f(x) = x^x = e^{x \ln x}\). Differentiate \(e^{x \ln x}\) using the chain rule to potentially get an improved estimate. Otherwise, we'll proceed numerically.

Step 5: Estimate Numerically

For a numerical approximation, choose a small \(h\), such as \(h = 0.01\), and calculate: \[\frac{(2.01)^{2.01} - 4}{0.01}.\]This will yield an approximation of the derivative at \(x = 2\).

Step 6: Compute the Approximation

Let's compute: \((2.01)^{2.01} \approx 4.0814\), so the expression is: \[\frac{4.0814 - 4}{0.01} = \frac{0.0814}{0.01} = 8.14.\]Thus, the estimated derivative \(f'(2)\) is approximately 8.14.

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique often used to simplify the process of differentiating complex functions. When dealing with functions like \( x^x \), the product and power rule of differentiation can get a bit tricky. However, by using logarithmic differentiation, the procedure becomes much clearer.

Here’s why it’s helpful:

• Transforming the Function: Using logarithms, we convert a function like \( x^x \) into a more manageable form: \( e^{x \ln x} \). This separation allows us to utilize properties of exponents, simplifying differentiation significantly.
• Applying Differentiation Rules: With the expression \( e^{x \ln x} \), we can apply the chain rule quickly. Differentiate the power \(x \ln x\) first, which involves taking the derivative of \(x\ln x\) using the product rule, resulting in \(\ln x + 1\).
• Resulting Derivative: The differentiated function is thus \( f'(x) = e^{x \ln x} (\ln x + 1) = x^x (\ln x + 1) \).

By implementing these steps, simplifying the derivative of a seemingly complex function becomes more attainable, especially in a conceptual framework.

Numerical Approximation

Often in mathematics, especially when dealing with derivatives, we resort to numerical approximations. This method is particularly useful when the algebraic manipulation involved is too complex or when an exact derivative is difficult to calculate.

• Choosing a small \(h\): To numerically approximate the derivative of a function like \(f(x) = x^x\) at \(x=2\), we pick a small \(h\) (often \(0.01\) or smaller). This gives a close approximation to the limit process in the definition of a derivative.
• Function Difference: We then compute \( f(2+h) - f(2)\). For example, if \(h=0.01\), \(f(2.01)\) gives us the approximate rate of change close to \(x=2\).
• Dividing by \(h\): The difference quotient \(\frac{f(2+h) - f(2)}{h}\) provides a numerical estimate of the derivative at \(x=2\), reflecting how the function changes in that small interval. This becomes our approximate derivative value.

With these steps, numerical approximation can lend insights into the behavior of derivatives when analytical methods are not readily applicable.

Chain Rule

The chain rule is a fundamental theorem in calculus providing a technique for finding the derivative of a composite function. When conductively paired with logarithmic differentiation, it makes differentiating complex functions simpler.

The essence of the chain rule is:

• Composite Functions: To differentiate functions like \( e^{x \ln x} \) (as seen in logarithmic differentiation), identify them as composite functions, which are functions within functions.
• Differentiate the Outer Function: Start by differentiating the outer function. For \(e^{x \ln x}\), this is differentiating \(e^{u}\), where \(u = x \ln x\), leading to \(e^u\cdot\frac{du}{dx}\).
• Inner Function Derivative: Then, find the derivative of the inner function \(x \ln x\) using the product rule, which results in \(\ln x + 1\).
• Combining Results: Multiply these results: \( e^{x \ln x} \cdot (\ln x + 1) \), rewriting it using the original expression \( x^x\). Therefore, the chain rule gives us the finished derivative \(f'(x) = x^x (\ln x + 1)\).

By applying the chain rule, complex derivative problems become structured, allowing for step-by-step problem-solving.