Problem 50 - Calculus for Biology and Medicine

Question

Differentiate the functions in Problems 1-52 with respect to the independent variable. $$ g(r)=2^{-3 \sin r} $$

Step-by-Step Solution

Answer

The derivative is \( g'(r) = -3 \ln 2 \cdot \cos r \cdot 2^{-3 \sin r} \).

Step 1: Recognize the Structure

The given function is \( g(r) = 2^{-3 \sin r} \). To differentiate it, recognize that \( 2^{-3 \sin r} \) can be rewritten as an exponential function using base \( e \) by expressing it in this form: \( 2^{u} = e^{u \ln 2} \). Thus, \( g(r) = e^{-3 \sin r \cdot \ln 2} \).

Step 2: Differentiate using Chain Rule

To differentiate \( e^{-3 \sin r \ln 2} \), use the chain rule. Let \( u = -3 \sin r \ln 2 \), then \( \frac{dg}{dr} = e^{u} \cdot \frac{du}{dr} \).

Step 3: Compute the Inner Derivative

Differentiate \( u = -3 \sin r \ln 2 \) with respect to \( r \). Use the derivative of \( \sin r \), which is \( \cos r \). Thus, \( \frac{du}{dr} = -3 \ln 2 \cdot \cos r \).

Step 4: Combine the Derivatives

Substitute the derivative back into the expression found in Step 2: \( \frac{dg}{dr} = e^{-3 \sin r \ln 2} \cdot (-3 \ln 2 \cdot \cos r) \). Simplify the expression: \( \frac{dg}{dr} = -3 \ln 2 \cdot \cos r \cdot 2^{-3 \sin r} \).

Step 5: Simplified Final Answer

Thus, the derivative of the given function with respect to \( r \) is \( g'(r) = -3 \ln 2 \cdot \cos r \cdot 2^{-3 \sin r} \).

Key Concepts

Chain Rule

The chain rule is a vital concept when dealing with composite functions. It helps us find derivatives of functions composed of other functions. For instance, in the function \( g(r) = 2^{-3 \sin r} \), we can break it down into the function \( -3 \sin r \) and the exponential function.

• When applying the chain rule, first identify the inner function, which in this example is \( u = -3 \sin r \).
• Then identify the outer function, here being the exponential function \( e^{u \ln 2} \).

The chain rule states that if a function \( g(x) = f(u(x)) \), then the derivative \( g'(x) = f'(u(x)) \cdot u'(x) \).
In our problem, it means that we differentiate \( e^{u \ln 2} \) with respect to \( u \) and multiply it by the derivative of \( u \) with respect to \( r \). This process allows us to handle complex functions by breaking them into simpler parts and differentiating each part separately.

Exponential Function

Exponential functions are a special type of mathematical function in the form \( a^{x} \), where \( a \) is a constant and \( x \) is the exponent variable. They often involve the natural base \( e \), which is approximately 2.718.

• In the given function \( g(r) = 2^{-3 \sin r} \), the base is 2 and the exponent is \(-3 \sin r\).
• We can express this using \( e \) by reconfiguring it as \( e^{-3 \sin r \cdot \ln 2} \).

This transformation is useful because derivatives of functions with base \( e \) have simpler rules. The derivative of \( e^{u} \) with respect to \( x \) is simply \( e^{u} \cdot \frac{du}{dx} \).
Recognizing this pattern in other bases, like when transforming \( 2^x \) to \( e^{x \ln 2} \), simplifies calculations and is a powerful tool in calculus.

Trigonometric Functions

Trigonometric functions, such as \( \sin \) and \( \cos \), are essential in calculus, especially when differentiating. In the function \( g(r) = 2^{-3 \sin r} \), \( \sin r \) serves as the inner function in our chain rule application.

• The derivative of \( \sin r \) is \( \cos r \).
• This derivative is crucial when applying the chain rule to differentiate the whole function.

Another important aspect is understanding how trigonometric functions change with respect to the angle \( r \). These functions are periodic and follow regular patterns, which assists in predicting their behavior after differentiation.
Using these derivatives effectively is essential in obtaining the outcome, which in this problem, contributes directly to calculating \( \frac{du}{dr} = -3 \ln 2 \cdot \cos r \). Mastering the differentiation of trigonometric functions allows us to handle more complex calculus problems efficiently.