Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$ $$ D_{x}\left[x^{\pi+1}+(\pi+1)^{x}\right] $$

Step-by-Step Solution

Answer

The derivative is \((\pi+1)x^{\pi} + (\pi+1)^x \ln(\pi+1)\)."

Step 1: Identify the Function Components

The given function to differentiate is \(f(x) = x^{\pi+1} + (\pi+1)^{x}\). It consists of two parts: \(x^{\pi+1}\) and \((\pi+1)^{x}\).

Step 2: Differentiate \(x^{\pi+1}\)

Use the power rule for differentiation, which states that \(\frac{d}{dx}x^n = nx^{n-1}\). Here, \(n = \pi+1\). So, the derivative of \(x^{\pi+1}\) is \((\pi+1)x^{\pi}\).

Step 3: Differentiate \((\pi+1)^{x}\)

Use the exponential differentiation rule for \(a^x\), which is \(\frac{d}{dx}a^x = a^x \ln a\), where \(a\) is a constant. For \((\pi+1)^x\), the derivative is \((\pi+1)^x \ln(\pi+1)\).

Step 4: Combine the Derivatives

Add the two derivatives from Step 2 and Step 3 to find the total derivative. The derivative of \(f(x) = x^{\pi+1} + (\pi+1)^{x}\) is \((\pi+1)x^{\pi} + (\pi+1)^{x} \ln(\pi+1)\).

Key Concepts

Power Rule in Calculus

The power rule is one of the most essential tools in calculus for finding derivatives. It allows us to quickly determine the rate of change for functions that involve powers of the variable. Simply put, if you have a function in the form of \(x^n\), its derivative will be \(nx^{n-1}\). Here’s why this is so useful:

• The power rule applies to any real number \(n\), simplifying complex expressions involving variables raised to a power.
• It's straightforward and quick, making it perfect for functions with polynomial terms.

In the exercise provided, we see this rule in action. We need to differentiate the term \(x^{\pi+1}\). According to the power rule, replacing \(n\) with \(\pi+1\), the derivative becomes \((\pi+1)x^{\pi}\). Each step follows the simple logic of bringing down the exponent and reducing the power by one.
The power rule greatly simplifies the process of differentiation, especially when dealing with polynomial expressions. It's part of the foundation for most calculus students, as its applications are vast and invaluable.

Exponential Functions

Exponential functions are a major category of functions that grow rapidly. They have the form \(a^x\), where \(a\) is a constant and \(x\) is the exponent. These functions are vital in many fields, from finance to population dynamics.
One key reason they are interesting is how they behave when we differentiate them. The derivative of an exponential function like \(a^x\) is unique in that it is proportional to itself. Specifically, the derivative is \(a^x \ln(a)\) because the base of the exponential function modifies the rate of growth.

• This rule helps us understand the growth rate of the function related to its natural log.
• The natural logarithm \(\ln(a)\) gives us a scaling factor that relates the function’s growth to its base.

In the exercise, we encounter the exponential function \((\pi+1)^x\). Its derivative is derived by applying this exponential rule, resulting in \((\pi+1)^x \ln(\pi+1)\). This finding highlights how the base affects the rate of change, a unique property of exponential functions.

Derivative Calculation

Calculating derivatives is a core skill in calculus. It helps us understand how functions change—essential for fields like physics, engineering, and economics. The process involves applying various differentiation rules to find the derivative of a given function.

• The calculations often start by identifying each part of the function that needs differentiation.
• Once identified, specific rules like the power rule and exponential rule are used to find the derivatives of these segments.

In our exercise, we applied two differentiation techniques: the power rule for \(x^{\pi+1}\) and the exponential differentiation rule for \((\pi+1)^x\). Combining these results gives us the derivative of the complete function. The total derivative, \((\pi+1)x^{\pi} + (\pi+1)^x \ln(\pi+1)\), is obtained by summing the derivatives of each term.
Understanding how to piece together these calculations is crucial. It requires knowledge of each rule and an ability to identify the type of function presented. By mastering these calculations, students can solve complex problems and predict how changes impact different scenarios.