Problem 48 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$ $$ D_{x}\left[2^{\left(e^{x}\right)}+\left(2^{e}\right)^{x}\right] $$

Step-by-Step Solution

Answer

The derivative is \( 2^{(e^x)} \ln(2) \cdot e^x + (2^e)^x \cdot e \cdot \ln(2) \)."

Step 1: Identify the Composite Functions

The function given is \( f(x) = 2^{(e^x)} + (2^e)^x \). This can be broken into separate functions where one is a power function with respect to \( e^x \), and the other is a power function with base \( 2^e \).

Step 2: Differentiate the First Term

The derivative of \( 2^{(e^x)} \) is found using the chain rule. The derivative is \( 2^{(e^x)} \ln(2) \cdot \frac{d}{dx}(e^x) = 2^{(e^x)} \ln(2) \cdot e^x \).

Step 3: Differentiate the Second Term

The derivative of \( (2^e)^x \) is straightforward: \( (2^e)^x \ln(2^e) \cdot \frac{d}{dx}(x) = (2^e)^x \cdot e \cdot \ln(2) \).

Step 4: Combine the Derivatives

Add the derivatives of both parts. Therefore, \( D_x[2^{(e^x)} + (2^e)^x] = 2^{(e^x)} \ln(2) \cdot e^x + (2^e)^x \cdot e \cdot \ln(2) \).

Key Concepts

Power Functions

Power functions are functions of the form \( f(x) = x^a \), where \( a \) is any constant exponent. In this exercise, we're dealing with power functions where exponents might not be integers, and they can even be functions themselves. This arises in expressions like \( 2^{(e^x)} \) and \((2^e)^x\).
Understanding power functions is essential because they are foundational in calculus. These expressions behave similarly to any other exponential form, but differentiation involves a more advanced method because of the variable exponents.

Chain Rule

The chain rule is a powerful tool in calculus used to differentiate composite functions. It allows us to "chain together" the derivatives of nested functions. For instance, when finding the derivative of \( 2^{(e^x)} \), the chain rule helps.
Here's how it works: if you have a function \( y \) expressed in terms of another function \( u \), like \( y = 2^u \) and \( u = e^x \), you would differentiate \( y \) with respect to \( u \) first, and then \( u \) with respect to \( x \).
This gives us the derivative: \[ \frac{dy}{dx} = 2^{(e^x)} \ln(2) \cdot e^x \].

• Differentiate the outer function: \( 2^u \) with respect to \( u \)
• Multiply by the derivative of the inner function: \( e^x \)

By using the chain rule, complex derivations become manageable.

Exponential Functions

Exponential functions are those that have a constant base and a variable exponent, like \( a^x \). They are crucial in mathematics because they model growth processes, among other things. In the given exercise, the function \( 2^{(e^x)} \) is particularly interesting because both its base and the exponent are constant and variable elements, respectively.
To differentiate \( a^x \), you utilize the natural logarithm since \( \frac{d}{dx}(a^x) = a^x \ln(a) \).
For instance, the differentiation of \( (2^e)^x \) forms part of the solution to the provided exercise. It's found as \( (2^e)^x \cdot \ln(2^e) \cdot \frac{d}{dx}(x)= (2^e)^x \cdot e \cdot \ln(2) \).

• Recognize the constant base: \( 2^e \)
• Differentiate using \( \ln \) of the base

Exponential functions are compelling due to their rapid growth nature.

Differentiation

Differentiation is the process of finding the derivative, or the rate of change, of a function. It is a fundamental operation in calculus, allowing us to find relations between dynamically changing quantities. In the task at hand, differentiation is applied to obtain the derivatives of \( 2^{(e^x)} \) and \( (2^e)^x \).
Differentiation involves applying rules and formulas like the power rule, chain rule, and the properties of exponential functions.
Since differentiation can handle combination and scaling, derivatives of functions like those given can be added directly:

• The first part gives: \( 2^{(e^x)} \ln(2) \cdot e^x \)
• The second part yields: \( (2^e)^x \cdot e \cdot \ln(2) \)

Adding these two gives the complete derivative of the function. This process forms the backbone of calculus, making it indispensable in mathematical analysis.

Composite Functions

Composite functions are created when one function is nested inside another, denoted as \( f(g(x)) \). These are encountered frequently in calculus and need special attention due to their layered nature. The function given \( f(x) = 2^{(e^x)} + (2^e)^x \) is a composite function where \( e^x \) is inside \( 2^u \).
Applying the chain rule is crucial here. You differentiate from the outer layer inward. For example:

• Differentiating \( 2^{(e^x)} \) necessitates differentiation of the outer base \( 2 \) and then the inner exponent \( e^x \)
• Through this process, we handle the layers by applying derivative operations inside out

Understanding composite functions and how to differentiate them ensures you can tackle a wide range of calculus problems with confidence.