Domanda del libro di testo
Evaluate the integrals in Exercises 31–78.
47. ∫(1/r)csc²(1+ln(r))dr
17
views
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
Non è quello che usi tu?Cambia libro di testoScegli un libro di testo diverso
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.P.27
In Exercises 25–30, use logarithmic differentiation to find the derivative of y with respect to the appropriate variable.
27. y = (((t+1)(t-1))/((t-2)(t+3)))^5, t>2
Guida verificata passo dopo passo1
Start by taking the natural logarithm of both sides of the equation to simplify the differentiation process: \(\ln y = \ln \left( \left( \frac{(t+1)(t-1)}{(t-2)(t+3)} \right)^5 \right)\).
Use the logarithm power rule to bring down the exponent: \(\ln y = 5 \ln \left( \frac{(t+1)(t-1)}{(t-2)(t+3)} \right)\).
Apply the logarithm quotient rule to separate the numerator and denominator inside the logarithm: \(\ln y = 5 \left( \ln (t+1) + \ln (t-1) - \ln (t-2) - \ln (t+3) \right)\).
Differentiate both sides with respect to \(t\), remembering that \(\frac{d}{dt} (\ln y) = \frac{1}{y} \frac{dy}{dt}\) by implicit differentiation: \(\frac{1}{y} \frac{dy}{dt} = 5 \left( \frac{1}{t+1} + \frac{1}{t-1} - \frac{1}{t-2} - \frac{1}{t+3} \right)\).
Finally, solve for \(\frac{dy}{dt}\) by multiplying both sides by \(y\): \(\frac{dy}{dt} = y \cdot 5 \left( \frac{1}{t+1} + \frac{1}{t-1} - \frac{1}{t-2} - \frac{1}{t+3} \right)\), and substitute back the original expression for \(y\) if desired.
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:
3mConcetti chiave
Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.
Logarithmic Differentiation
Logarithmic differentiation is a technique used to differentiate functions that are products, quotients, or powers of other functions. By taking the natural logarithm of both sides, the differentiation process simplifies using properties of logarithms, especially useful for complicated expressions raised to a power.
Video consigliato:
06:30
Logarithmic Differentiation
Properties of Logarithms
The properties of logarithms, such as log(ab) = log a + log b, log(a/b) = log a - log b, and log(a^n) = n log a, allow complex expressions to be broken down into simpler sums and differences. These properties are essential in logarithmic differentiation to simplify the function before differentiating.
Video consigliato:
05:36Change of Base Property
Chain Rule
The chain rule is a fundamental differentiation rule used when differentiating composite functions. In logarithmic differentiation, after taking the logarithm, the derivative of the outer function (logarithm) is combined with the derivative of the inner function, enabling the calculation of the original function's derivative.
Video consigliato:
05:02Intro to the Chain Rule
Pratica correlata
Domanda del libro di testo
In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
13. y = (x+2)^(x+2)
37
views
Domanda del libro di testo
In Exercises 125–128 solve the differential equation.
125. dy/dx = √y cos(√y)
6
views
Domanda del libro di testo
112. True, or false? Give reasons for your answers.
c. ln x = o(x+1)
25
views
Domanda del libro di testo
118. A particle is traveling upward and to the right along the curve y=ln(x). Its x-coordinate is increasing at the rate (dx/dt)=√x m/sec. At what rate is the y-coordinate changing at the point (e², 2)?
27
views
Domanda del libro di testo
111. True, or false? Give reasons for your answers.
e. arctan x = O(1)
19
views