Problem 46 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. $$ g(s)=\ln \left(\sin ^{2}(3 s)\right) $$

Step-by-Step Solution

Answer

The derivative of \( g(s) = \ln(\sin^2(3s)) \) is \( 3 \csc(3s) \).

Step 1: Identify the Outer Function

The function to differentiate is given by \[ g(s) = \ln(\sin^2(3s)) \]The outer function is the natural logarithm, \( \ln(x) \). We'll use the chain rule to differentiate this function, \( u = \sin^2(3s) \).

Step 2: Differentiate the Outer Function

First, differentiate the outer function \( \ln(u) \) with respect to \( u \):\[ \frac{d}{du}[\ln(u)] = \frac{1}{u} \]Thus, for our function, the derivative is \[ \frac{1}{\sin^2(3s)} \cdot \frac{d}{ds}[\sin^2(3s)] \]

Step 3: Differentiate the Inner Function

We now need to differentiate the inner function \( \sin^2(3s) \) with respect to \( s \). Using the chain rule where \( v = \sin(3s) \), we differentiate \( v^2 \):\[ \frac{d}{dv}[v^2] = 2v \cdot \frac{d}{ds}[\sin(3s)] \]

Step 4: Differentiate Further Inside

Differentiate \( \sin(3s) \) with respect to \( s \):\[ \frac{d}{ds}[\sin(3s)] = 3 \cos(3s) \]Substitute back:\[ 2 \sin(3s) \cdot 3 \cos(3s) = 6 \sin(3s) \cos(3s) \]

Step 5: Combine Results

Substitute the result from differentiating the inner function into the derivative of the outer function:\[ \frac{d}{ds}[g(s)] = \frac{1}{\sin^2(3s)} \cdot 6 \sin(3s) \cos(3s) \]Simplify:\[ \frac{6 \sin(3s) \cos(3s)}{\sin^2(3s)} = 6 \cot(3s)\cos(3s) \]The final result uses the identity \( \frac{\cos(3s)}{\sin(3s)} = \cot(3s) \).

Step 6: Finalize the Simplification

The derivative can be simplified further using the double angle identity:\[ 6 \cdot \frac{1}{\sin(3s)} \cdot \cos(3s) = 3 \cdot \frac{2\cos(3s)}{\sin(3s)} = 3 \cdot \cot (3s) \cdot 2\cos(3s) = 3 \cdot \sin(6s) \]The simplest form is:\[ 3 \csc(3s) \] as \( \sin(6s) = 2\sin(3s)\cos(3s) \).

Key Concepts

Chain Rule

The chain rule is a vital concept in calculus that helps us differentiate complex functions. It allows us to "chain" together the derivatives of different functions. When you have a function composed of layers, like an onion, you peel back each layer using derivatives.

Why is this important? Because many real-world problems don't come wrapped in simple, straightforward functions. The chain rule lets us tackle the more complicated ones too.

• When a function is nested, such as \( g(s) = \ln(\sin^2(3s)) \), it has an "outer" function \( \ln(x) \) and an "inner" function \( \sin^2(3s) \).
• Start with the outer layer: differentiate the natural logarithm function with respect to its argument.
• Next, dive into the inner function and apply differentiation rules again.

In this exercise, we first took the derivative of the natural logarithm, leaving us with \( \frac{1}{\sin^2(3s)} \). Then, using the chain rule again, we handled the inner function by differentiating \( \sin^2(3s) \).

Trigonometric Functions

Trigonometric functions play a central role in calculus, especially when dealing with periodic phenomena. They describe the relationships between angles and side lengths in right triangles.

For derivatives, memorize:

• Derivative of \( \sin(x) \) is \( \cos(x) \).
• Derivative of \( \cos(x) \) is \( -\sin(x) \).
• When inside composite functions, use the chain rule to find the rate of change.

In this problem, \( \sin^2(3s) \) is decomposed using trigonometric identities and derivatives. After differentiating \( \sin(3s) \), which gives us \( 3\cos(3s) \), the derivative is pieced together using known trigonometric properties.

Understanding how to manage derivatives involving trigonometric functions using these basic rules can greatly simplify complex calculus problems.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a function that's inverse to the exponential function, \( e^x \). It's a powerful tool in calculus due to its unique derivative properties.

Key properties of \( \ln(x) \):

• \( \ln(1) = 0 \)
• \( \ln(e) = 1 \)
• Its derivative is \( \frac{1}{x} \).

When combined with the chain rule, the natural logarithm's derivative properties enable the differentiation of nested functions, such as in this exercise: \( g(s)=\ln(\sin^2(3s)) \). We first identified the differentiable piece inside the logarithm and applied the derivative of \( \ln(u) \) as \( \frac{1}{u} \).

This approach allows tackling complex mathematical relationships, rendering them into manageable pieces for further analysis and solution.