In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
45. y=cos(x-arccos(x))

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

Apply the chain rule to differentiate \(y = \cos(u)\) where \(u = x - \arccos(x)\). The derivative of \(\cos(u)\) with respect to \(x\) is \(-\sin(u) \cdot \frac{du}{dx}\).

Find \(\frac{du}{dx}\) where \(u = x - \arccos(x)\). Differentiate each term separately: the derivative of \(x\) with respect to \(x\) is 1, and the derivative of \(\arccos(x)\) with respect to \(x\) is \(-\frac{1}{\sqrt{1 - x^2}}\).

Combine the derivatives to get \(\frac{du}{dx} = 1 - \left(-\frac{1}{\sqrt{1 - x^2}}\right) = 1 + \frac{1}{\sqrt{1 - x^2}}\).

Substitute back into the chain rule formula: \(\frac{dy}{dx} = -\sin\left(x - \arccos(x)\right) \cdot \left(1 + \frac{1}{\sqrt{1 - x^2}}\right)\).

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주요 개념

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Derivative of Trigonometric Functions

Understanding how to differentiate trigonometric functions like cosine is essential. The derivative of cos(u) with respect to x is -sin(u) times the derivative of u, applying the chain rule to handle composite functions.

Chain Rule

The chain rule is used to differentiate composite functions. It states that the derivative of f(g(x)) is f'(g(x)) multiplied by g'(x). This is crucial when differentiating expressions like cos(x - arccos(x)).

Derivative of Inverse Trigonometric Functions

Knowing the derivative of inverse trig functions, such as arccos(x), is important. The derivative of arccos(x) is -1 divided by the square root of (1 - x^2), which is needed when differentiating terms involving arccos(x).

In Exercises 21–48, find the derivative of y with respect to the appropriate variable.45. y=cos(x-arccos(x))