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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
Tutti i libri di testoHass 15th EditionCh. 7 - Transcendental FunctionsProblema 7.3.119
Capitolo 7, Problema 7.3.119

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.
119. y = (sin x)ˣ

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1
Recognize that the function is of the form \(y = (\sin x)^x\), which is a variable base raised to a variable exponent. This is a perfect candidate for logarithmic differentiation.
Take the natural logarithm of both sides to simplify the expression: \(\ln y = \ln \left( (\sin x)^x \right)\).
Use the logarithm power rule to bring the exponent down: \(\ln y = x \cdot \ln (\sin x)\).
Differentiate both sides with respect to \(x\). Remember to use implicit differentiation on the left side and the product rule on the right side: \(\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left( x \ln (\sin x) \right)\).
Apply the product rule to the right side: \(\frac{d}{dx} \left( x \ln (\sin x) \right) = 1 \cdot \ln (\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x\). Then multiply both sides by \(y\) to solve for \(\frac{dy}{dx}\).

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Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions where the variable appears both in the base and the exponent, such as y = (sin x)^x. By taking the natural logarithm of both sides, the expression simplifies to a product, making it easier to apply differentiation rules.
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Logarithmic Differentiation

Chain Rule

The chain rule is a fundamental differentiation rule used when differentiating composite functions. In this problem, it helps differentiate expressions like ln(sin x) and sin x, which are nested inside other functions, by multiplying the derivative of the outer function by the derivative of the inner function.
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Intro to the Chain Rule

Implicit Differentiation

Implicit differentiation involves differentiating both sides of an equation with respect to the independent variable, treating y as a function of x. After taking logarithms, y is expressed implicitly, and this method helps find dy/dx by differentiating terms involving y and x together.
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Finding The Implicit Derivative
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