In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
21. y = z arcsec(z) - √(z² - 1), z>1

검증된 단계별 안내

Identify the function to differentiate: \(y = z \cdot \operatorname{arcsec}(z) - \sqrt{z^{2} - 1}\), where \(z > 1\).

Recall the derivative of \(\operatorname{arcsec}(z)\) with respect to \(z\): \(\frac{d}{dz} \operatorname{arcsec}(z) = \frac{1}{|z| \sqrt{z^{2} - 1}}\). Since \(z > 1\), \(|z| = z\).

Apply the product rule to the first term \(z \cdot \operatorname{arcsec}(z)\): \(\frac{d}{dz} [z \cdot \operatorname{arcsec}(z)] = \operatorname{arcsec}(z) \cdot \frac{d}{dz} z + z \cdot \frac{d}{dz} \operatorname{arcsec}(z)\).

Differentiate the second term \(- \sqrt{z^{2} - 1}\) using the chain rule: rewrite as \(-(z^{2} - 1)^{1/2}\) and find its derivative.

Combine the derivatives from the product rule and the chain rule to write the full expression for \(\frac{dy}{dz}\).

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주요 개념

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

Derivative of Inverse Trigonometric Functions

Inverse trigonometric functions like arcsec(x) have specific derivative formulas. For arcsec(x), the derivative is 1 / (|x|√(x² - 1)) when |x| > 1. Understanding this formula is essential to differentiate expressions involving arcsec.

Product Rule for Differentiation

The product rule is used to differentiate products of two functions. If y = u(x)v(x), then y' = u'v + uv'. Applying this rule correctly is crucial when differentiating terms like z * arcsec(z).

Chain Rule and Differentiation of Composite Functions

The chain rule helps differentiate composite functions, such as √(z² - 1). It states that the derivative of f(g(x)) is f'(g(x)) * g'(x). This rule is necessary to handle the square root of an expression involving z.

In Exercises 1–24, find the derivative of y with respect to the appropriate variable.21. y = z arcsec(z) - √(z² - 1), z>1