In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
29. y=arcsec(1/t), 0<t<1

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Identify the function given: \(y = \arcsec\left(\frac{1}{t}\right)\), where \(0 < t < 1\).

Recall the derivative formula for \(y = \arcsec(u)\) with respect to \(t\): \(\frac{dy}{dt} = \frac{1}{|u| \sqrt{u^2 - 1}} \cdot \frac{du}{dt}\), where \(u\) is a function of \(t\).

Set \(u = \frac{1}{t}\) and compute its derivative with respect to \(t\): \(\frac{du}{dt} = \frac{d}{dt} \left(\frac{1}{t}\right) = -\frac{1}{t^2}\).

Substitute \(u\) and \(\frac{du}{dt}\) back into the derivative formula: \(\frac{dy}{dt} = \frac{1}{\left| \frac{1}{t} \right| \sqrt{\left(\frac{1}{t}\right)^2 - 1}} \cdot \left(-\frac{1}{t^2}\right)\).

Simplify the expression step-by-step, paying attention to the absolute value and the square root, to express \(\frac{dy}{dt}\) in terms of \(t\).

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주요 개념

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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)). Understanding these formulas is essential to differentiate functions involving inverse trig expressions.

Chain Rule

The chain rule is used to differentiate composite functions. When y = arcsec(1/t), you first differentiate arcsec(u) with respect to u, then multiply by the derivative of u = 1/t. This rule allows handling nested functions systematically.

Domain Considerations for arcsec(x)

The domain of arcsec(x) is |x| ≥ 1, but here the argument is 1/t with 0 < t < 1, so 1/t > 1, which fits the domain. Recognizing domain restrictions ensures the function and its derivative are valid in the given interval.

In Exercises 21–48, find the derivative of y with respect to the appropriate variable.29. y=arcsec(1/t), 0<t<1