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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.2.70
Chapter 7, Problem 7.2.70

In Exercises 57–70, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.
70. y = ∛(x(x+1)(x-2)/(x²+1)(2x+3))

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Start by rewriting the function to make logarithmic differentiation easier. Let \( y = \sqrt[3]{\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}} \). This can be expressed as \( y = \left( \frac{x(x+1)(x-2)}{(x^2+1)(2x+3)} \right)^{\frac{1}{3}} \).
Take the natural logarithm of both sides to simplify the differentiation process: \( \ln y = \frac{1}{3} \ln \left( \frac{x(x+1)(x-2)}{(x^2+1)(2x+3)} \right) \).
Use the logarithm property for quotients and products to expand the right side: \( \ln y = \frac{1}{3} \left[ \ln x + \ln (x+1) + \ln (x-2) - \ln (x^2+1) - \ln (2x+3) \right] \).
Differentiate both sides with respect to \( x \). Remember that \( \frac{d}{dx} (\ln y) = \frac{1}{y} \frac{dy}{dx} \) by implicit differentiation. So, \( \frac{1}{y} \frac{dy}{dx} = \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x+1} + \frac{1}{x-2} - \frac{2x}{x^2+1} - \frac{2}{2x+3} \right) \).
Finally, solve for \( \frac{dy}{dx} \) by multiplying both sides by \( y \), and then substitute back the original expression for \( y \) to express the derivative in terms of \( x \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate complex functions, especially those involving products, quotients, or powers. By taking the natural logarithm of both sides, the function is simplified into sums and differences, making differentiation easier. This method is particularly useful when the function is a product or quotient of multiple expressions.
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Logarithmic Differentiation

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 expressions into simpler parts. These rules help transform products into sums and quotients into differences, which simplifies the differentiation process after taking the logarithm of the function.
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Change of Base Property

Chain Rule

The chain rule is a fundamental differentiation technique used when dealing with composite functions. After applying logarithmic differentiation, the resulting expression often involves functions inside other functions, requiring the chain rule to differentiate correctly. It states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.
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