In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
15. y = sin⁻¹√(1-u²), 0<u<1

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Identify the function given: \(y = \sin^{-1}\left(\sqrt{1 - u^2}\right)\), where \(0 < u < 1\).

Recall the derivative formula for the inverse sine function: if \(y = \sin^{-1}(x)\), then \(\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}\).

Set the inner function as \(g(u) = \sqrt{1 - u^2}\) and apply the chain rule: \(\frac{dy}{du} = \frac{1}{\sqrt{1 - (g(u))^2}} \cdot g'(u)\).

Calculate \(g'(u)\) by differentiating \(g(u) = (1 - u^2)^{1/2}\) using the power rule and chain rule: \(g'(u) = \frac{1}{2}(1 - u^2)^{-1/2} \cdot (-2u)\).

Substitute \(g(u)\) and \(g'(u)\) back into the expression for \(\frac{dy}{du}\) and simplify the resulting expression step-by-step.

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주요 개념

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Inverse Trigonometric Functions

Inverse trigonometric functions, like sin⁻¹(x), are the inverses of the standard trig functions and return an angle whose trigonometric value is x. Understanding their domains and ranges is essential for correctly differentiating expressions involving them.

Chain Rule

The chain rule is a differentiation technique used when a function is composed of another function. 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 Inverse Sine Function

The derivative of y = sin⁻¹(x) with respect to x is 1/√(1 - x²), valid for |x| < 1. This formula is crucial when differentiating expressions involving inverse sine, especially when combined with the chain rule.

In Exercises 1–24, find the derivative of y with respect to the appropriate variable.15. y = sin⁻¹√(1-u²), 0<u<1