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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
Capitolo 7, Problema 7.2.64

In Exercises 57–70, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.
64. y = 1/(t(t+1)(t+2))

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Start by expressing the function clearly: \(y = \frac{1}{t(t+1)(t+2)}\). This can also be written as \(y = [t(t+1)(t+2)]^{-1}\) to prepare for logarithmic differentiation.
Take the natural logarithm of both sides to simplify the product inside the logarithm: \(\ln y = \ln \left( [t(t+1)(t+2)]^{-1} \right)\).
Use the logarithm power rule to bring down the exponent: \(\ln y = -1 \cdot \ln [t(t+1)(t+2)]\).
Apply the logarithm product rule to expand the right side: \(\ln y = - (\ln t + \ln (t+1) + \ln (t+2))\).
Differentiate both sides with respect to \(t\). Remember that \(\frac{d}{dt} (\ln y) = \frac{1}{y} \frac{dy}{dt}\) by the chain rule, and differentiate each logarithmic term on the right side accordingly.

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Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions that are products, quotients, or powers of variables by taking the natural logarithm of both sides. This simplifies the differentiation process by converting multiplication into addition and division into subtraction, making it easier to handle complex expressions.
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Properties of Logarithms

The properties of logarithms, such as log(ab) = log a + log b and log(a/b) = log a - log b, allow us to break down complicated products and quotients into sums and differences. These properties are essential in logarithmic differentiation to simplify the function before differentiating.
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Implicit Differentiation

Implicit differentiation involves differentiating both sides of an equation with respect to the independent variable, treating the dependent variable as a function of that variable. In logarithmic differentiation, after taking logs, implicit differentiation is used to find dy/dx by differentiating the transformed equation.
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