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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Key Concepts

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

Derivative of the Natural Logarithm Function

The derivative of ln(u), where u is a differentiable function of a variable, is (1/u) times the derivative of u. This rule allows us to differentiate logarithmic expressions by first identifying the inner function and then applying the chain rule.

Product Rule for Differentiation

When differentiating a product of two functions, the product rule states that the derivative is the first function times the derivative of the second plus the second function times the derivative of the first. This is essential when the argument of the logarithm is a product, like 3t and e^(-t).

Derivative of the Exponential Function

The derivative of e^f(t), where f(t) is a function of t, is e^f(t) times the derivative of f(t). Recognizing this helps differentiate terms like e^(-t), which appear inside the logarithm's argument.

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