In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
37. y=s√(1-s²) + arccos(s)

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Identify the function to differentiate: \(y = s \sqrt{1 - s^2} + \arccos(s)\), where \(y\) is expressed in terms of \(s\).

Rewrite the square root term for easier differentiation: \(s \sqrt{1 - s^2} = s (1 - s^2)^{1/2}\).

Apply the product rule to differentiate \(s (1 - s^2)^{1/2}\). Recall the product rule: \(\frac{d}{ds}[u v] = u' v + u v'\), where \(u = s\) and \(v = (1 - s^2)^{1/2}\).

Differentiate \(\arccos(s)\) using the chain rule. The derivative of \(\arccos(s)\) with respect to \(s\) is \(-\frac{1}{\sqrt{1 - s^2}}\).

Combine the derivatives from the product rule and the derivative of \(\arccos(s)\) to write the full expression for \(\frac{dy}{ds}\).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Derivative of Composite Functions

This involves applying the chain rule when differentiating functions composed of other functions, such as √(1 - s²). The chain rule 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 Trigonometric Functions

The derivative of arccos(s) with respect to s is -1/√(1 - s²). Understanding the derivatives of inverse trig functions is essential for differentiating expressions involving arccos or arcsin.

Product Rule for Differentiation

When differentiating a product of two functions, such as s and √(1 - s²), the product rule is used. It states that the derivative of f(s)g(s) is f'(s)g(s) + f(s)g'(s), combining the derivatives of each function appropriately.

In Exercises 21–48, find the derivative of y with respect to the appropriate variable.37. y=s√(1-s²) + arccos(s)