In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.
y = ln(3te^(-t))

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Identify the function given: $y = \ln(3te^{-t})$. Notice that the argument of the logarithm is a product of two functions: \$3t$ and $e^{-t}$.

Use the logarithm property to simplify the expression before differentiating: $\ln(3te^{-t}) = \ln(3t) + \ln(e^{-t})$.

Further simplify using the logarithm of an exponential: $\ln(e^{-t}) = -t$, so the function becomes $y = \ln(3t) - t$.

Differentiate each term separately with respect to $t$. For $\ln(3t)$, use the chain rule: $\frac{d}{dt}[\ln(3t)] = \frac{1}{3t} \times 3 = \frac{1}{t}$. For $-t$, the derivative is $-1$.

Combine the derivatives to write the final derivative: $\frac{dy}{dt} = \frac{1}{t} - 1$.

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