In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
17. y = ln(arccos(x))

Verified step by step guidance

1. Identify the function given: \(y = \ln(\arccos(x))\). We need to find \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\).
2. Recall the chain rule for derivatives: if \(y = \ln(u)\), then \(\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}\). Here, \(u = \arccos(x)\).
3. Find the derivative of the inner function \(u = \arccos(x)\). The derivative is \(\frac{du}{dx} = -\frac{1}{\sqrt{1 - x^2}}\).
4. Apply the chain rule by substituting \(u\) and \(\frac{du}{dx}\) into the formula: \(\frac{dy}{dx} = \frac{1}{\arccos(x)} \cdot \left(-\frac{1}{\sqrt{1 - x^2}}\right)\).
5. Simplify the expression to write the derivative as \(\frac{dy}{dx} = -\frac{1}{\arccos(x) \sqrt{1 - x^2}}\).

Key Concepts

Chain Rule

The chain rule is a fundamental differentiation technique used to find the derivative of composite functions. It states that the derivative of f(g(x)) is f'(g(x)) multiplied by g'(x). In this problem, since y = ln(arccos(x)) is a composition of ln(u) and u = arccos(x), the chain rule is essential.

Derivative of the Natural Logarithm Function

The derivative of ln(u), where u is a differentiable function of x, is 1/u times the derivative of u. This rule helps differentiate logarithmic functions by simplifying the process to the derivative of the inner function divided by the original function.

Derivative of the Arccosine Function

The derivative of arccos(x) with respect to x is -1 divided by the square root of (1 - x^2). This formula is crucial for differentiating inverse trigonometric functions and is needed to find the derivative of the inner function in the given problem.