Question

In Exercises 59–86, find the derivative of y with respect to the given independent variable.77. y = log₃(((x + 1)/(x − 1))^(ln 3))

Accepted Answer

Recognize that the function is given by $$y = \log_{3} \left( \left( \frac{x+1}{x-1} \$$right)^{\ln 3}$$ \right). $$The goal is to find $$\frac{dy}{dx}. $$Use the logarithm power rule to simplify the expression inside the logarithm: $$\log_{3} \left( a$$^{b}$$ \right) = b \cdot \log_{3}(a). $$Applying this, rewrite $$y as$$ $$y = (\ln 3) \cdot \log_{3} \left( \frac{x+1}{x-1} \right). $$Recall the change of base formula for logarithms: $$\log_{a}(b) = \frac{\ln b}{\ln a}. $$Use this to rewrite $$\log_{3} \left( \frac{x+1}{x-1} \right) as$$ $$\frac{\ln \left( \frac{x+1}{x-1} \right)}{\ln 3}. $$Substitute this back into $$y to $$get $$y = (\ln 3) \cdot \frac{\ln \left( \frac{x+1}{x-1} \right)}{\ln 3}. $$Notice that $$\ln 3$$ cancels out, simplifying $$y to$$ $$y = \ln \left( \frac{x+1}{x-1} \right). $$Now differentiate $$y = \ln \left( \frac{x+1}{x-1} \right)$$ using the chain rule. The derivative of $$\ln u$$ with respect to $$x is$$ $$\frac{1}{u} \cdot \frac{du}{dx}. $$Here, $$u = \frac{x+1}{x-1}. $$Find $$\frac{du}{dx}$$ using the quotient rule and then write $$\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}.$$

In Exercises 59–86, find the derivative of y with respect to the given independent variable.
77. y = log₃(((x + 1)/(x − 1))^(ln 3))

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주요 개념

Logarithmic Functions and Change of Base

Chain Rule

Logarithmic Differentiation