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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.6.25
Chapter 7, Problem 7.6.25

In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
25. y=arcsec(2s+1)

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1
Identify the function given: \(y = \arcsec(2s + 1)\). We need to find \(\frac{dy}{ds}\), the derivative of \(y\) with respect to \(s\).
Recall the derivative formula for \(y = \arcsec(u)\), where \(u\) is a function of \(s\): \(\frac{dy}{ds} = \frac{1}{|u| \sqrt{u^2 - 1}} \cdot \frac{du}{ds}\).
Set \(u = 2s + 1\). Compute the derivative of \(u\) with respect to \(s\): \(\frac{du}{ds} = 2\).
Substitute \(u\) and \(\frac{du}{ds}\) into the derivative formula: \(\frac{dy}{ds} = \frac{1}{|2s + 1| \sqrt{(2s + 1)^2 - 1}} \cdot 2\).
Simplify the expression inside the square root and write the final derivative expression in terms of \(s\), leaving it in simplified radical form without calculating the numeric value.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative of Inverse Trigonometric Functions

Inverse trigonometric functions like arcsec(x) have specific derivative formulas. Understanding these formulas is essential to differentiate functions involving inverse trig expressions correctly.
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Derivatives of Other Inverse Trigonometric Functions

Chain Rule

The chain rule is used to differentiate composite functions. When the argument of arcsec is a function of x, such as 2s+1, the derivative of the outer function is multiplied by the derivative of the inner function.
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Intro to the Chain Rule

Domain and Differentiability of arcsec(x)

The function arcsec(x) is defined for |x| ≥ 1, and its derivative involves expressions with sqrt(x^2 - 1). Recognizing the domain restrictions and the form of the derivative helps avoid errors in differentiation.
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Finding the Domain and Range of a Graph
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