Problem 48 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 \( f'(x) = 2^{e^x} e^x \ln(2) + (2^e)^x e \ln(2) \).

Step 1: Understand the Function

The given function is \( f(x) = 2^{e^x} + (2^e)^x \). It consists of two parts: \( 2^{e^x} \) and \( (2^e)^x \). We need to differentiate both parts with respect to \(x\).

Step 2: Differentiate the First Part

Let's start with the first part \( 2^{e^x} \). Use the chain rule for derivatives. First, the derivative of \(2^u\) with respect to \(u\) is \(2^u \ln(2)\). Here \(u = e^x\), which gives the additional derivative \(e^x\cdot e^x\) when deriving \(e^x\) with respect to \(x\). So the derivative of the first part is:\[ \frac{d}{dx} [2^{e^x}] = 2^{e^x} \ln(2) \cdot e^x.\]

Step 3: Differentiate the Second Part

For the second part \( (2^e)^x \), apply the power rule for exponential functions. The derivative of \(a^x\) with respect to \(x\) is \(a^x \ln(a)\). Thus, for \((2^e)^x\), it becomes:\[ \frac{d}{dx} [(2^e)^x] = (2^e)^x \ln(2^e). \]

Step 4: Simplify the Second Part's Derivative

Notice \( \ln(2^e) = e \ln(2) \). Hence, the derivative of the second part simplifies to:\[ (2^e)^x \cdot e \cdot \ln(2). \]

Step 5: Combine the Derivatives

Combine the derivatives of both parts found in Steps 2 and 4:\[ f'(x) = 2^{e^x} \cdot e^x \cdot \ln(2) + (2^e)^x \cdot e \cdot \ln(2). \]

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used for finding the derivative of compositions of functions. When you have a function within another function, the chain rule helps you differentiate it by considering how each function affects the derivative.
In our original exercise, we applied the chain rule to differentiate the expression \(2^{e^x}\). Essentially, this involves two steps:

• Differentiate the outer function: When you have \( 2^{e^x} \), treat \( e^x \) as a single variable \( u \) (where \( u = e^x \)) and take the derivative of \( 2^u \) which is \( 2^u \ln(2) \).
• Differentiate the inner function: The inner function here is \( e^x \), and its derivative with respect to \( x \) is \( e^x \).

By multiplying these derivatives together, you employ the chain rule to find the complete derivative of the composite function. It helps in simplifying complex derivatives by breaking them into more manageable parts.

Exponential Functions

Exponential functions are mathematical functions of the form \( a^x \), where \( a \) is a positive constant and \( x \) is a variable. These functions are crucial because they model exponential growth or decay in natural phenomena.

In the given problem, we dealt with two exponential terms:

• \( 2^{e^x} \) represents an exponential function where the exponent itself is an exponential expression.
• \( (2^e)^x \) where the base is an exponent with the constant \( e \).

Differentiating such functions includes using specific rules adapted to handle their exponential nature. Notice how we used the natural logarithm in our solutions' steps. This logarithmic relationship simplifies the differentiation process and underlines how fundamental the notion of the natural log is in handling exponents.

Power Rule

The power rule is a straightforward and direct method to find derivatives of power functions, specifically those in the form \( x^n \). It states that if you have \( f(x) = x^n \), its derivative \( f'(x) \) would be \( n \cdot x^{n-1} \).

However, in the context of exponential functions, a related but more generalized rule applies. For functions of the form \( a^x \), the differentiation rule becomes:

• \( \frac{d}{dx} [a^x] = a^x \ln(a) \)

In our exercise, it was applied to the part \((2^e)^x\). Using the formula, the derivative is computed as \( (2^e)^x \ln(2^e) \). This showcases how a similar principle to the power rule simplifies differentiating exponential functions effortlessly. It’s important to memorize these rules, as they make solving calculus problems much more efficient and manageable.