Problem 66 - Calculus for Biology and Medicine

Question

Use logarithmic differentiation to find the first derivative of the given functions. $$ f(x)=(2 x)^{2 x} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = (2x)^{2x} \cdot (2 \ln(2x) + 2) \).

Step 1: Take the Natural Logarithm of Both Sides

Given the function \( f(x) = (2x)^{2x} \), take the natural logarithm of both sides to simplify differentiation: \( \ln(f(x)) = \ln((2x)^{2x}) \).

Step 2: Simplify the Logarithmic Expression

Use the property of logarithms that \( \ln(a^b) = b \cdot \ln(a) \): \( \ln(f(x)) = 2x \cdot \ln(2x) \).

Step 3: Differentiate Both Sides with Respect to x

Differentiate both sides of the equation. For the left side: \( \frac{d}{dx}[\ln(f(x))] = \frac{1}{f(x)} \cdot f'(x) \). For the right side: Use the product rule \( \frac{d}{dx}[2x \cdot \ln(2x)] = 2 \cdot \ln(2x) + 2x \cdot \frac{d}{dx}(\ln(2x)) \).

Step 4: Differentiate \( \ln(2x) \)

To differentiate \( \ln(2x) \), use the chain rule: \( \frac{d}{dx}[\ln(2x)] = \frac{1}{2x} \cdot 2 = \frac{1}{x} \).

Step 5: Combine Derivatives

Insert the differentiation result from Step 4 into the expression from Step 3: \( \frac{d}{dx}[2x \cdot \ln(2x)] = 2 \ln(2x) + 2 \).

Step 6: Solve for \( f'(x) \)

Using the differentiated form \( \frac{1}{f(x)} \cdot f'(x) = 2 \ln(2x) + 2 \), solve for \( f'(x) \): \( f'(x) = f(x) \cdot (2 \ln(2x) + 2) \).

Step 7: Substitute Original Function

Substitute \( f(x) = (2x)^{2x} \) back into the derivative expression: \( f'(x) = (2x)^{2x} \cdot (2 \ln(2x) + 2) \).

Key Concepts

Understanding Derivative Calculation

When dealing with calculus, derivatives are fundamental in understanding how functions change. The derivative of a function gives the rate at which the function's value changes with respect to changes in its input.
Logarithmic differentiation is a unique technique used for finding derivatives of complicated functions, particularly those that have variables in both the base and the exponent, like the function given in the exercise: \(f(x) = (2x)^{2x}\).
Here, the natural log simplifies this differentiation process. By taking the natural logarithm of both sides, complex exponentiation becomes manageable algebra through properties of logarithms.

• Use the natural log to transform powers and products into sums and differences.
• Remember that \( \ln(a^b) = b \ln(a) \), which is crucial for simplifying the logarithmic expression before differentiation.

Exploring the Product Rule

The product rule is a differentiation technique used when dealing with functions that are the product of two or more functions, which was applied in our solution during differentiation.
When a function \( h(x) \) is the product of two differentiable functions \( u(x) \) and \( v(x) \), the derivative is found using:

• \( \frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \).

This rule was used when differentiating \( 2x \cdot \ln(2x) \). Here, \( u(x) = 2x \) and \( v(x) = \ln(2x) \). Knowing how to apply the product rule increases efficiency when handling multiplication inside functions.

Mastering the Chain Rule

In calculus, the chain rule helps differentiate composite functions. It's crucial for handling operations within embedded functions, like \( \ln(2x) \), as in our example.
If \( g(x) \) is a function nested within \( f(x) \), the chain rule tells us that:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).

For \( \ln(2x) \), here \( f(x) = \ln(x) \) and \( g(x) = 2x \). The chain rule simplifies differentiation by breaking down complexity into a product of their derivatives. Applying this gives us \( \frac{1}{x} \), as shown during logarithmic differentiation in the original exercise.

Differentiation Techniques Overview

Logarithmic differentiation, product rule, and chain rule are just a few techniques in calculus to untangle the complexity of functions. They provide systematic ways to find derivatives for challenging functions.

• **Logarithmic Differentiation** simplifies expressions with variable exponents.
• **Product Rule** helps deal with products of separate differentiable functions.
• **Chain Rule** unveils the derivative of composite functions.

Together, these strategies enable handling differentiation with ease, turning intimidating calculus problems into approachable tasks. Practice and familiarity with these rules facilitate tackling any derivative calculation efficiently.