Problem 52 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the indicated derivative. \(D_{x} x^{\left(2^{x}\right)}\)

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = x^{2^x} \cdot (\ln x \cdot 2^x \cdot \ln 2 + \frac{2^x}{x})\)

Step 1: Understand the Problem

We need to find the derivative of the function \(f(x) = x^{(2^x)}\). This involves understanding how to differentiate expressions where the exponent varies with x.

Step 2: Rewrite Using Exponential and Logarithm

To handle the varying exponent, rewrite the expression using logarithmic differentiation. Let \(y = x^{2^x}\). Taking the natural logarithm on both sides gives \(\ln y = (2^x) \cdot \ln x\).

Step 3: Differentiate Implicitly

Differentiate both sides with respect to \(x\). Recall that \(\frac{d}{dx}(\ln y) = \frac{1}{y}\cdot \frac{dy}{dx}\). So, we have \(\frac{1}{y}\cdot \frac{dy}{dx} = \ln x \cdot \frac{d}{dx}(2^x) + 2^x \cdot \frac{d}{dx}(\ln x)\).

Step 4: Evaluate Derivatives on Right Side

Next, differentiate the terms on the right side. We have \(\frac{d}{dx}(2^x) = 2^x \cdot \ln 2\), and \(\frac{d}{dx}(\ln x) = \frac{1}{x}\). Substituting these gives: \(\frac{1}{y}\cdot \frac{dy}{dx} = \ln x \cdot 2^x \cdot \ln 2 + \frac{2^x}{x}\).

Step 5: Solve for \(\frac{dy}{dx}\)

Multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\): \(\frac{dy}{dx} = y \cdot (\ln x \cdot 2^x \cdot \ln 2 + \frac{2^x}{x})\). Remember, \(y = x^{2^x}\), so substitute back to get \(\frac{dy}{dx} = x^{2^x} \cdot (\ln x \cdot 2^x \cdot \ln 2 + \frac{2^x}{x})\).

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful tool that helps us differentiate functions where the exponent is a function of the variable, such as exponentials with variable bases. This technique is especially useful in our example where we need to differentiate the function \(f(x) = x^{2^x}\).

Here's why we use logarithmic differentiation:
- It simplifies the differentiation process by taking the natural log of both sides, which linearizes the problem.
- This method makes it easier to differentiate complex functions involving variable exponents.

To apply logarithmic differentiation:
1. Start with the original function: Let's say \(y = x^{2^x}\).
2. Take the natural logarithm on both sides, resulting in: \(\ln y = (2^x) \cdot \ln x\).
3. Differentiate both sides with respect to \(x\).

This transformation takes advantage of logarithmic properties and makes the differentiation problem much more manageable.

Exponential Functions

Exponential functions appear frequently in calculus, and their derivatives often involve exponential terms. In our function \(x^{2^x}\), the base is \(x\) and the exponent is \(2^x\), a rapidly growing expression.

Understanding how to differentiate exponential functions is key:
- The derivative of \(a^x\) where \(a\) is a constant is \(a^x \cdot \ln a\). This shows how exponentially growing bases have derivatives that grow alongside them.
- In our case, we have an exponential term \(2^x\) in the exponent, making things a bit more complex.

Specific to our problem:
- Differentiate this exponential term using the rule: \(\frac{d}{dx}(2^x) = 2^x \cdot \ln 2\).
This differentiation allows us to handle the exponential component of the derivative correctly and incorporate it back into our expression.

Implicit Differentiation

Implicit differentiation is used when we have a function defined implicitly rather than explicitly, and we need to find its derivative. It comes in handy when direct differentiation is challenging.

This technique worked for our function. We began with:
- \(y = x^{2^x}\)
Taking the natural log on both sides allowed us to create an implicit relationship: \(\ln y = (2^x) \cdot \ln x\).

Implicit differentiation steps:

• Differentiate both sides with respect to \(x\).
• On the left, this gives us \(\frac{1}{y} \cdot \frac{dy}{dx}\). We need to apply the chain rule here because \(y\) is a function of \(x\).
• On the right, apply product and chain rules wherever applicable, as in differentiating \((2^x)\) and \(\ln x\).

Using implicit differentiation, we solve for \(\frac{dy}{dx}\), the derivative of the original function, leaving us with a more tractable solution for complex problems.