Problem 49 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(g(x)=2^{2 \cos x}\)

Step-by-Step Solution

Answer

The derivative is \(-2 \cdot 2^{2 \cos x} \ln(2) \sin x\).

Step 1: Identify the Function

Recognize that the function to differentiate is given as \( g(x) = 2^{2 \cos x} \). This is an exponential function with a base of 2 and a power that is a function of \( x \).

Step 2: Use the Chain Rule

To differentiate \( g(x) = 2^{2 \cos x} \), we set \( u = 2 \cos x \). Therefore, \( g(x) = 2^u \). The derivative of \( a^u \) with respect to \( x \) is \( a^u \ln(a) \frac{du}{dx} \).

Step 3: Differentiate the Exponential Function

Apply the formula for differentiating exponential functions: the derivative of \( 2^u \) with respect to \( x \) is \( 2^u \ln(2) \frac{du}{dx} \). Here, \( a = 2 \), so \( \ln(a) = \ln(2) \).

Step 4: Find the Derivative of the Exponent

Differentiate the exponent \( u = 2 \cos x \) with respect to \( x \). The derivative of \( \cos x \) is \( -\sin x \). Therefore, \( \frac{du}{dx} = 2 \cdot -\sin x = -2 \sin x \).

Step 5: Apply the Chain Rule and Simplify

Combine the results from Steps 3 and 4. The derivative \( g'(x) \) becomes \( 2^{2 \cos x} \ln(2) \cdot (-2 \sin x) \), which simplifies to \( -2^{2 \cos x} \cdot 2 \ln(2) \sin x \).

Step 6: Finalize the Derivative Expression

Simplify the expression if needed. The derivative \( g'(x) \) can be rewritten as \( -2 \cdot 2^{2 \cos x} \ln(2) \sin x \).

Key Concepts

Understanding the Chain Rule

The chain rule is an essential tool when differentiating composite functions, where one function is nested inside another. In simpler terms, it's a way of finding the derivative of a function that depends on another function. Here's how it works:

• Identify the outer and inner functions. In our example function, \( g(x) = 2^{2 \cos x} \), the outer function is the exponential \( 2^u \) and the inner function is the trigonometric part \( u = 2 \cos x \).
• Differentiation requires you to compute the derivative of the outer function while keeping the inner function intact, and then multiply it with the derivative of the inner function.

The chain rule can be expressed as \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \). In this context, this means taking the derivative of the exponential function and then multiplying it by the derivative of the cosine function.
By recognizing how functions are nested, you can effectively differentiate more complex expressions using the chain rule.

Exploring Exponential Functions

Exponential functions, characterized by having a constant base raised to a variable exponent, are unique and appear frequently in mathematical modeling. The function in our problem is an example, \( g(x) = 2^{2 \cos x} \), with base 2.

• Exponential functions are known for their constant growth rates, making them suitable for modeling continuous growth processes.
• When differentiating exponential functions, the key formula is \( \frac{d}{dx}[a^u] = a^u \ln(a) \frac{du}{dx} \). Here, \( a \) is the base, and \( u \) is the exponent, which is typically a function of \( x \).
• In our case, you differentiate the whole expression by first applying the chain rule to manage the exponent, then applying the standard exponential differentiation rule.

Grasping this differentiation approach allows you to tackle problems involving exponential growth and decay effectively.

The Role of Trigonometric Differentiation

Trigonometric differentiation plays a vital role when dealing with functions containing trigonometric expressions, such as sine and cosine. These functions exhibit periodic behavior and are integral to various real-world applications.

• When you have a term like \( \cos x \) in an expression, its derivative is \( -\sin x \). This derivative is crucial when applying the chain rule in compounded functions.
• For example, in the expression \( u = 2 \cos x \), the differentiation yields \( \frac{du}{dx} = -2 \sin x \), which we use in the chain rule. Multiplying this result with other derivatives helps you reach the final form, \( g'(x) \).
• Understanding these rules allows you to calculate the rate of change in trigonometric contexts, enhancing your ability to solve complex calculus problems.

Through practice, recognizing the derivatives of trigonometric functions and how to apply them will become second nature, allowing smoother problem-solving.