Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problema 7.6.130

Capitolo 7, Problema 7.6.130

130. Use the identity arccot(u)=π/2 - arctan(u) to derive the formula for the derivative of arccot(u) in Table 7.4 from the formula for the derivative of arctan(u).

Guida verificata passo dopo passo

Recall the given identity: \(\arccot(u) = \frac{\pi}{2} - \arctan(u)\).

Differentiate both sides of the equation with respect to \(x\). Since \(\frac{\pi}{2}\) is a constant, its derivative is zero, so we have: \(\frac{d}{dx} [\arccot(u)] = \frac{d}{dx} \left[ \frac{\pi}{2} - \arctan(u) \right] = - \frac{d}{dx} [\arctan(u)]\).

Use the chain rule to differentiate \(\arctan(u)\): \(\frac{d}{dx} [\arctan(u)] = \frac{1}{1 + u^2} \cdot \frac{du}{dx}\).

Substitute this result back into the derivative of \(\arccot(u)\): \(\frac{d}{dx} [\arccot(u)] = - \frac{1}{1 + u^2} \cdot \frac{du}{dx}\).

Thus, the derivative formula for \(\arccot(u)\) is derived as \(\frac{d}{dx} [\arccot(u)] = - \frac{1}{1 + u^2} \cdot \frac{du}{dx}\), matching the formula in Table 7.4.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Inverse Trigonometric Functions

Inverse trigonometric functions, like arctan and arccot, are the inverses of the tangent and cotangent functions, respectively. They return the angle whose tangent or cotangent is a given number. Understanding their definitions and domains is essential for manipulating and differentiating these functions.

Video consigliato:

06:35

Derivatives of Other Inverse Trigonometric Functions

Derivative of arctan(u)

The derivative of arctan(u) with respect to x is given by (1 / (1 + u^2)) times the derivative of u. This formula is fundamental because it provides a direct way to differentiate arctan functions, which can then be used to find derivatives of related inverse trig functions through identities.

Video consigliato:

Percorso guidato

05:44

Derivatives

Using Identities to Derive Derivatives

By expressing one inverse trig function in terms of another using identities, such as arccot(u) = π/2 - arctan(u), we can apply known derivatives and the chain rule to find new derivatives. This method simplifies the process by leveraging existing formulas rather than differentiating from first principles.

Video consigliato:

4:42

Solve Trig Equations Using Identity Substitutions

Pratica correlata

Domanda del libro di testo

Solve the initial value problems in Exercises 55–58.

57. d²y/dx² = 2e^(−x),y(0) = 1,y′(0) = 0

views

Domanda del libro di testo

25. First-order chemical reactions In some chemical reactions, the rate at which the amount of a substance changes with time is proportional to the amount present. For the change of δ-gluconolactone into gluconic acid, for example,

dy/dt = -0.6y

when t is measured in hours. If there are 100 grams of δ-gluconolactone present when t=0, how many grams will be left after the first hour?

views

Domanda del libro di testo

Evaluate the integrals in Exercises 91–102.

102. ∫(from -1/3 to 1/√3)(cos(arctan 3x))/(1+9x²) dx

views

Domanda del libro di testo

Theory and Applications

L’Hôpital’s Rule does not help with the limits in Exercises 69–76.

Try it—you just keep on cycling. Find the limits some other way.

71. lim (x → (π/2)⁻) sec x / tan x

views

Domanda del libro di testo

In Exercises 115–126, use logarithmic differentiation or the method in Example 6 to find the derivative of y with respect to the given independent variable.

122. y = (ln x)^(ln x)

views

Domanda del libro di testo

In Exercises 57–70, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.

66. y = θsin(θ)/√(sec(θ))

views