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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.6.39

In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
39. y=arctan√(x²-1) + arccsc(x), x>1

Guida verificata passo dopo passo
1
Identify the function to differentiate: \(y = \arctan\left(\sqrt{x^{2} - 1}\right) + \arccsc(x)\), where \(x > 1\).
Recall the derivative formulas: For \(y = \arctan(u)\), \(\frac{dy}{dx} = \frac{1}{1 + u^{2}} \cdot \frac{du}{dx}\); for \(y = \arccsc(x)\), \(\frac{dy}{dx} = -\frac{1}{|x| \sqrt{x^{2} - 1}}\).
Differentiate the first term: Let \(u = \sqrt{x^{2} - 1} = (x^{2} - 1)^{1/2}\). Compute \(\frac{du}{dx}\) using the chain rule: \(\frac{du}{dx} = \frac{1}{2}(x^{2} - 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^{2} - 1}}\).
Apply the derivative formula for \(\arctan(u)\): \(\frac{d}{dx} \arctan\left(\sqrt{x^{2} - 1}\right) = \frac{1}{1 + (\sqrt{x^{2} - 1})^{2}} \cdot \frac{x}{\sqrt{x^{2} - 1}}\).
Differentiate the second term \(\arccsc(x)\) using its derivative formula: \(\frac{d}{dx} \arccsc(x) = -\frac{1}{|x| \sqrt{x^{2} - 1}}\). Since \(x > 1\), \(|x| = x\).

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