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

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Identify the function given: \(y = \ln(\arccos(x))\). We need to find \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\).

Recall the chain rule for derivatives: if \(y = \ln(u)\), then \(\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}\). Here, \(u = \arccos(x)\).

Find the derivative of the inner function \(u = \arccos(x)\). The derivative is \(\frac{du}{dx} = -\frac{1}{\sqrt{1 - x^2}}\).

Apply the chain rule by substituting \(u\) and \(\frac{du}{dx}\) into the formula: \(\frac{dy}{dx} = \frac{1}{\arccos(x)} \cdot \left(-\frac{1}{\sqrt{1 - x^2}}\right)\).

Simplify the expression to write the derivative as \(\frac{dy}{dx} = -\frac{1}{\arccos(x) \sqrt{1 - x^2}}\).

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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.

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Intro to the Chain Rule

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.

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Derivative of the Natural Logarithmic 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.

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