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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.6.27

In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
27. y=arccsc(x²+1)

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1
Identify the function to differentiate: \(y = \arccsc(x^{2} + 1)\), where the argument of the arccsc function is \(u = x^{2} + 1\).
Recall the derivative formula for \(y = \arccsc(u)\) with respect to \(x\): \(\frac{dy}{dx} = -\frac{1}{|u| \sqrt{u^{2} - 1}} \cdot \frac{du}{dx}\).
Compute the derivative of the inner function \(u = x^{2} + 1\): \(\frac{du}{dx} = 2x\).
Substitute \(u\) and \(\frac{du}{dx}\) into the derivative formula: \(\frac{dy}{dx} = -\frac{1}{|x^{2} + 1| \sqrt{(x^{2} + 1)^{2} - 1}} \cdot 2x\).
Simplify the expression under the square root if possible, and write the final derivative expression in terms of \(x\).

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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 arccsc(x) have specific derivative formulas. For arccsc(u), the derivative is -u' / (|u|√(u² - 1)), where u is a function of x. Understanding these formulas is essential to differentiate expressions involving inverse trig functions.
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Derivatives of Other Inverse Trigonometric Functions

Chain Rule

The chain rule is used to differentiate composite functions. When y = f(g(x)), the derivative y' = f'(g(x)) * g'(x). In this problem, since the argument of arccsc is x² + 1, the chain rule helps differentiate the inner function x² + 1.
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Intro to the Chain Rule

Simplification and Domain Considerations

After differentiation, simplifying the expression is important for clarity. Also, recognizing the domain restrictions of arccsc and its derivative, such as the argument being outside the interval (-1,1), ensures the derivative is valid and correctly interpreted.
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