Question

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\u003e2

Accepted Answer

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.

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

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Key Concepts

Logarithmic Differentiation

Properties of Logarithms

Chain Rule

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