Problem 339 - Calculus Volume 1

Question

For the following exercises, find \(f^{\prime}(x)\) for each function. $$f(x)=x^{\pi} \cdot \pi^{x}$$

Step-by-Step Solution

Answer

The derivative is \(f^{\prime}(x) = \pi^x (\pi x^{\pi-1} + x^{\pi} \ln(\pi))\).

Step 1: Identify the Problem Type

The problem asks us to find the derivative \(f^{\prime}(x)\) of the function given by \(f(x) = x^{\pi} \cdot \pi^{x}\). This requires the application of the product rule and derivative formulas for power and exponential functions.

Step 2: Apply the Product Rule

According to the product rule for differentiation, if \(f(x) = u(x) \cdot v(x)\), then the derivative \(f^{\prime}(x) = u^{\prime}(x) \cdot v(x) + u(x) \cdot v^{\prime}(x)\). Here, let \(u(x) = x^{\pi}\) and \(v(x) = \pi^x\).

Step 3: Differentiate \(u(x) = x^{\pi}\)

The derivative of \(u(x) = x^{\pi}\) is \(u^{\prime}(x) = \pi x^{\pi - 1}\). This follows from the power rule, where the derivative of \(x^n\) is \(nx^{n-1}\).

Step 4: Differentiate \(v(x) = \pi^x\)

The derivative of the exponential function \(v(x) = \pi^x\) is \(v^{\prime}(x) = \pi^x \ln(\pi)\). This is because the derivative of \(a^x\) is \(a^x \ln(a)\), where \(a\) is a constant.

Step 5: Combine Results Using Product Rule

Now apply the product rule: \[f^{\prime}(x) = u^{\prime}(x) \cdot v(x) + u(x) \cdot v^{\prime}(x) = \pi x^{\pi-1} \cdot \pi^x + x^{\pi} \cdot \pi^x \ln(\pi)\]

Step 6: Simplify the Expression

Factor out the common term \(\pi^x\) from both parts of the expression: \[f^{\prime}(x) = \pi^x (\pi x^{\pi-1} + x^{\pi} \ln(\pi))\] This results in the final simplified derivative.

Key Concepts

Product Rule

When you have a function that's the product of two sub-functions, the product rule helps you find the derivative.
To use this rule, you write one function as \( u(x) \) and the other as \( v(x) \). The rule states:

• The derivative \( f^{\prime}(x) \) is calculated as \( u^{\prime}(x) \cdot v(x) + u(x) \cdot v^{\prime}(x) \).
• This means you differentiate each function separately and then combine them using addition.

Using the product rule in our exercise, we identify \( u(x) = x^{\pi} \) and \( v(x) = \pi^x \). You compute each derivative separately, \( u^{\prime}(x) \) and \( v^{\prime}(x) \), and apply the rule to find \( f^{\prime}(x) \).
This method is crucial for handling any product of functions.

Power Rule

The power rule is a straightforward method for differentiating functions of the form \( x^n \), where \( n \) is any real number.
The general idea is simple:

• If you have a function \( x^n \), its derivative is \( n \cdot x^{n-1} \).
• This means you bring down the exponent as a coefficient and subtract one from the exponent.

In the example, for \( u(x) = x^{\pi} \), we apply the power rule to find that \( u^{\prime}(x) = \pi x^{\pi-1} \).
Remember, the power rule is a powerful tool for quickly solving derivatives of polynomials and similar functions.

Exponential Functions

For exponential functions like \( \pi^x \), the derivative calculation takes a unique yet straightforward approach. These functions have the form \( a^x \), where \( a \) is a constant:

• The derivative is \( a^x \ln(a) \).
• This relies on the natural logarithm, \( \ln \), which is key to handling the exponential base.

In our solution, \( v(x) = \pi^x \) is differentiated to get \( v^{\prime}(x) = \pi^x \ln(\pi) \).
Understanding how to differentiate an exponential function is essential for problems involving exponential growth or decay, as it often shows up in real-world scenarios.
With practice, applying this rule becomes second nature.