91Ó°ÊÓ

The chain rule is a fundamental technique in calculus used for differentiating composite functions. These are functions that are composed of two or more functions. Think of it like peeling layers of an onion — you handle one layer at a time. In this exercise, the function to differentiate is a natural logarithm of another function, which itself is a composite of trigonometric components.

To apply the chain rule, you consider the outer function first, which in this case is the natural log function, \( y = \ln(u) \). If \( u \) is some function of \( x \), you first find \( \frac{d}{dx}[\ln(u)] = \frac{1}{u} \). Then you multiply it by the derivative of the inside function \( \frac{du}{dx} \).
Key components of the chain rule:

• Identify the inner function \( u(x) \).
• Differentiated the outer function \( \ln(u) \).
• Multiply the derivative of the outer by the derivative of the inner function.

This structured approach is crucial for correctly differentiating complex expressions.

Differentiating trigonometric functions is an essential skill in calculus and comes up frequently, especially in composite functions. The exercise involves taking the derivative of trigonometric components: \( \csc x \) and \( \cot x \).

For reference, here are common trigonometric derivatives:

• \( \frac{d}{dx}[\sin x] = \cos x \)
• \( \frac{d}{dx}[\cos x] = -\sin x \)
• \( \frac{d}{dx}[\tan x] = \sec^2 x \)
• \( \frac{d}{dx}[\cot x] = -\csc^2 x \)
• \( \frac{d}{dx}[\sec x] = \sec x \tan x \)
• \( \frac{d}{dx}[\csc x] = -\csc x \cot x \)

The exercise heavily relies on understanding how these functions behave when differentiated.

In our exercise:- \( \frac{d}{dx}[\csc x] = -\csc x \cot x \)- \( \frac{d}{dx}[-\cot x] = \csc^2 x \)These derivatives are essential to applying the chain rule properly and obtaining a simplified result.

The differentiation of natural logarithms, written as \( \ln(x) \), plays a significant role when dealing with composite functions. The derivative formula \( \frac{d}{dx} [\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx} \) is central to this exercise.

This formula tells us that to differentiate \( \ln(u) \), we need to:

• Let \( u = \text{expression inside the ln} \) — that's our inner function \( u(x) \).
• Derivative of \( \ln(u) \) involves creating a fraction with \( 1/u \) as the base and multiplying it by the derivative of the inner function \( \frac{du}{dx} \).

This approach ensures that we handle both the natural logarithm function and its inner complexity simultaneously.

Combining the chain rule with natural logarithm differentiation allows us to simplify and analyze functions involving logarithmic and trigonometric elements together, resulting in a complete solution to the given problem.