Problem 43 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. $$ f(x)=\ln (\sin x) $$

Step-by-Step Solution

Answer

The derivative of \( f(x) = \ln(\sin x) \) is \( \cot x \).

Step 1: Identify the Outer and Inner Functions

In the function \( f(x)=\ln (\sin x) \), the outer function is the natural logarithm \( \ln(u) \) and the inner function is \( u = \sin x \). This indicates we will use the chain rule for differentiation.

Step 2: Differentiate the Outer Function

First, differentiate the outer function \( \ln(u) \) with respect to \( u \), which gives \( \frac{d}{du}\ln(u) = \frac{1}{u} \).

Step 3: Differentiate the Inner Function

Now, differentiate the inner function \( \sin x \) with respect to \( x \), which gives \( \frac{d}{dx}\sin x = \cos x \).

Step 4: Apply the Chain Rule

Using the chain rule, the derivative of \( f(x)\) is the product of the derivatives found in Steps 2 and 3: \[ \frac{df}{dx} = \frac{1}{\sin x} \cdot \cos x \].

Step 5: Simplify the Expression

The expression can be simplified: \[ \frac{df}{dx} = \frac{\cos x}{\sin x} = \cot x \]. The derivative of \( f(x) = \ln(\sin x) \) with respect to \( x \) is \( \cot x \).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus for finding the derivative of a composition of functions. To put it simply, if one function is nested inside another, the chain rule helps you differentiate the combined function.

The process involves:

• Identifying the "outer" and "inner" functions.
• Differentiating these functions separately.
• Multiplying the derivative of the outer function by the derivative of the inner function.

In our example, the function is \( f(x) = \ln(\sin x) \). Here, the outer function is \( \ln(u) \) where \( u = \sin x \).
The inner function is \( \sin x \).

Once you differentiate the outer function \( \ln(u) \), which is \( \frac{1}{u} \), you multiply it by the derivative of the inner function \( \sin x \), which is \( \cos x \). This gives us the derivative: \( \frac{\cos x}{\sin x} = \cot x \).
The chain rule connects these steps and is essential for tackling more complex differentiation problems.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm with base \( e \), where \( e \approx 2.718 \). It's incredibly useful in calculus due to its unique properties.

One of its key features is how it interacts with differentiation. For instance, the derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). This property is handy when differentiating functions such as \( \ln(\sin x) \).

Here's what you need to remember about natural logarithms:

• They turn multiplicative relationships into additive ones, making complex problems simpler.
• The derivative behavior \( \frac{1}{u} \) simplifies differentiation involving product and composite functions.

In our differentiation here, it enables simplification when dealing with the logarithm of trigonometric expressions, leading to a clearer understanding and simpler derivative.

Trigonometric Functions

Trigonometric functions are the cornerstone of many mathematical calculations, especially those involving periodic phenomena. The basic ones include sine, cosine, and tangent.

Here's a quick breakdown of their role in differentiation:

• The sine function \( \sin x \)
• The cosine function \( \cos x \)
• The tangent function \( \tan x \)

When differentiating trigonometric functions, specific rules apply:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \( -\sin x \).
• The derivative of \( \tan x \) is \( \sec^2 x \).

In our example, the function \( \sin x \) nested inside \( \ln(\sin x) \) leads to a chain of operations involving these trigonometric derivatives. Mastery of these derivatives allows for tackling a wide range of differentiation problems efficiently.