Domanda del libro di testo
Evaluate the integrals in Exercises 31–78.
55. ∫(from -2 to -1)e^(-(x+1)) dx
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Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 7, Problema 7.PE.21
In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
21. y = z arcsec(z) - √(z² - 1), z>1
Guida verificata passo dopo passo1
Identify the function to differentiate: \(y = z \cdot \operatorname{arcsec}(z) - \sqrt{z^{2} - 1}\), where \(z > 1\).
Recall the derivative of \(\operatorname{arcsec}(z)\) with respect to \(z\): \(\frac{d}{dz} \operatorname{arcsec}(z) = \frac{1}{|z| \sqrt{z^{2} - 1}}\). Since \(z > 1\), \(|z| = z\).
Apply the product rule to the first term \(z \cdot \operatorname{arcsec}(z)\): \(\frac{d}{dz} [z \cdot \operatorname{arcsec}(z)] = \operatorname{arcsec}(z) \cdot \frac{d}{dz} z + z \cdot \frac{d}{dz} \operatorname{arcsec}(z)\).
Differentiate the second term \(- \sqrt{z^{2} - 1}\) using the chain rule: rewrite as \(-(z^{2} - 1)^{1/2}\) and find its derivative.
Combine the derivatives from the product rule and the chain rule to write the full expression for \(\frac{dy}{dz}\).
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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)) when |x| > 1. Understanding this formula is essential to differentiate expressions involving arcsec.
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06:35
Derivatives of Other Inverse Trigonometric Functions
Product Rule for Differentiation
The product rule is used to differentiate products of two functions. If y = u(x)v(x), then y' = u'v + uv'. Applying this rule correctly is crucial when differentiating terms like z * arcsec(z).
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05:18The Product Rule
Chain Rule and Differentiation of Composite Functions
The chain rule helps differentiate composite functions, such as √(z² - 1). It states that the derivative of f(g(x)) is f'(g(x)) * g'(x). This rule is necessary to handle the square root of an expression involving z.
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03:58The Chain Rule for 3+ Functions
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