Domanda del libro di testo
Each of Exercises 19–24 gives a formula for a function y=f(x) and shows the graphs of f and f^(-1). Find a formula for f^(-1) in each case.
f(x)=x²+1, x≥0
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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 7.6.45
In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
45. y=cos(x-arccos(x))
Guida verificata passo dopo passo1
Identify the function given: \(y = \cos\left(x - \arccos(x)\right)\). We need to find \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\).
Apply the chain rule to differentiate \(y = \cos(u)\) where \(u = x - \arccos(x)\). The derivative of \(\cos(u)\) with respect to \(x\) is \(-\sin(u) \cdot \frac{du}{dx}\).
Find \(\frac{du}{dx}\) where \(u = x - \arccos(x)\). Differentiate each term separately: the derivative of \(x\) with respect to \(x\) is 1, and the derivative of \(\arccos(x)\) with respect to \(x\) is \(-\frac{1}{\sqrt{1 - x^2}}\).
Combine the derivatives to get \(\frac{du}{dx} = 1 - \left(-\frac{1}{\sqrt{1 - x^2}}\right) = 1 + \frac{1}{\sqrt{1 - x^2}}\).
Substitute back into the chain rule formula: \(\frac{dy}{dx} = -\sin\left(x - \arccos(x)\right) \cdot \left(1 + \frac{1}{\sqrt{1 - x^2}}\right)\).
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Derivative of Trigonometric Functions
Understanding how to differentiate trigonometric functions like cosine is essential. The derivative of cos(u) with respect to x is -sin(u) times the derivative of u, applying the chain rule to handle composite functions.
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Derivatives of Other Inverse Trigonometric Functions
Chain Rule
The chain rule is used to differentiate composite functions. It states that the derivative of f(g(x)) is f'(g(x)) multiplied by g'(x). This is crucial when differentiating expressions like cos(x - arccos(x)).
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Derivative of Inverse Trigonometric Functions
Knowing the derivative of inverse trig functions, such as arccos(x), is important. The derivative of arccos(x) is -1 divided by the square root of (1 - x^2), which is needed when differentiating terms involving arccos(x).
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06:35Derivatives of Other Inverse Trigonometric Functions
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