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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
Tutti i libri di testoHass 15th EditionCh. 7 - Transcendental FunctionsProblema 7.7.31
Capitolo 7, Problema 7.7.31

In Exercises 25–36, find the derivative of y with respect to the appropriate variable.
31. y = cos⁻¹(x) - x sech⁻¹(x)

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Identify the function to differentiate: \(y = \cos^{-1}(x) - x \ \text{sech}^{-1}(x)\), where \(\cos^{-1}(x)\) is the inverse cosine function and \(\text{sech}^{-1}(x)\) is the inverse hyperbolic secant function.
Recall the derivative formulas for the inverse functions involved: - For \(y = \cos^{-1}(x)\), the derivative is \(\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}}\). - For \(y = \text{sech}^{-1}(x)\), the derivative is \(\frac{dy}{dx} = -\frac{1}{x \sqrt{1 - x^2}}\) (valid for \(0 < x < 1\)).
Apply the product rule to the term \(x \ \text{sech}^{-1}(x)\): If \(u = x\) and \(v = \text{sech}^{-1}(x)\), then \(\frac{d}{dx}(uv) = u'v + uv'\). Calculate \(u' = 1\) and use the derivative of \(v\) from the previous step.
Write the derivative of \(y\) as: \(\frac{dy}{dx} = \frac{d}{dx} \left( \cos^{-1}(x) \right) - \frac{d}{dx} \left( x \ \text{sech}^{-1}(x) \right)\), which becomes \(\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} - \left( 1 \cdot \text{sech}^{-1}(x) + x \cdot \left(-\frac{1}{x \sqrt{1 - x^2}}\right) \right)\).
Simplify the expression by distributing the negative sign and combining like terms, keeping the derivative in terms of \(x\) without evaluating the final numeric value.

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Derivative of Inverse Trigonometric Functions

Inverse trigonometric functions like arccos(x) have specific derivative formulas. For example, the derivative of arccos(x) is -1 / √(1 - x²), valid for |x| < 1. Understanding these derivatives is essential for differentiating expressions involving inverse trig functions.
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Derivatives of Other Inverse Trigonometric Functions

Derivative of Inverse Hyperbolic Functions

Inverse hyperbolic functions such as sech⁻¹(x) have distinct derivative rules. The derivative of sech⁻¹(x) is -1 / (|x|√(1 - x²)) for appropriate domains. Recognizing and applying these formulas is crucial when differentiating terms involving inverse hyperbolic functions.
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Derivatives of Other Inverse Trigonometric Functions

Product Rule for Differentiation

When differentiating a product of two functions, use the product rule: (fg)' = f'g + fg'. This rule is necessary here because the term x sech⁻¹(x) is a product of x and sech⁻¹(x). Applying the product rule correctly ensures accurate differentiation of such expressions.
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The Product Rule
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Since the hyperbolic functions can be expressed in terms of exponential functions, it is possible to express the inverse hyperbolic functions in terms of logarithms, as shown in the following table.

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coth⁻¹x = (1/2)ln((x+1)/(x-1)), |x| > 1

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