In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
43. y=√(arcsin x)

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Identify the function given: \(y = \sqrt{\arcsin x}\). This can be rewritten as \(y = (\arcsin x)^{\frac{1}{2}}\) to make differentiation easier.

Apply the chain rule for differentiation. The chain rule states that if \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). Here, \(f(u) = u^{\frac{1}{2}}\) and \(g(x) = \arcsin x\).

Differentiate the outer function \(f(u) = u^{\frac{1}{2}}\) with respect to \(u\): \(f'(u) = \frac{1}{2} u^{-\frac{1}{2}}\).

Differentiate the inner function \(g(x) = \arcsin x\) with respect to \(x\): \(g'(x) = \frac{1}{\sqrt{1 - x^2}}\).

Combine the results using the chain rule: \(\frac{dy}{dx} = \frac{1}{2} (\arcsin x)^{-\frac{1}{2}} \cdot \frac{1}{\sqrt{1 - x^2}}\).

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주요 개념

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Derivative of Inverse Trigonometric Functions

The derivative of inverse trigonometric functions like arcsin(x) is essential for this problem. Specifically, the derivative of arcsin(x) with respect to x is 1 / √(1 - x²), which helps in differentiating composite functions involving arcsin.

Chain Rule

The chain rule is used to differentiate composite functions, such as y = √(arcsin x). It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

Derivative of Square Root Functions

The derivative of a square root function, like √u, can be found using the power rule by rewriting it as u^(1/2). Its derivative is (1/2)u^(-1/2) times the derivative of u, which is crucial when differentiating y = √(arcsin x).

In Exercises 21–48, find the derivative of y with respect to the appropriate variable.43. y=√(arcsin x)