Problem 56 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\ln \left(3 \theta e^{-\theta}\right)$$

Step-by-Step Solution

Answer

The derivative is \( \frac{1}{\theta} - 1 \).

Step 1: Understanding the Function

The given function is \( y = \ln(3\theta e^{-\theta}) \). This function is a composite function involving a logarithmic function and an exponential function.

Step 2: Apply the Logarithm Property

Use the property \( \ln(ab) = \ln(a) + \ln(b) \) to simplify the function: \( y = \ln(3\theta) + \ln(e^{-\theta}) \).

Step 3: Simplify Using Logarithm of Exponent

Recognize that \( \ln(e^{-\theta}) = -\theta \). Thus, the function becomes \( y = \ln(3\theta) - \theta \).

Step 4: Separate Terms for Differentiation

Identify that the expression is now \( y = \ln(3) + \ln(\theta) - \theta \). We can ignore \( \ln(3) \) in differentiation since it's a constant.

Step 5: Differentiate Each Term

Differentiate \( \ln(\theta) \) to get \( \frac{d}{d\theta}[ \ln(\theta) ] = \frac{1}{\theta} \). Differentiate \( -\theta \) to get \( \frac{d}{d\theta}[-\theta] = -1 \).

Step 6: Combine the Derivatives

Combine the derivatives to find \( \frac{dy}{d\theta} = \frac{1}{\theta} - 1 \).

Key Concepts

Composite Function

A composite function is where one function is applied to the result of another function. In our exercise, the function is a combination of a logarithmic and an exponential function. This is a slightly more advanced concept because you need to consider how the functions interact with each other. When taking the derivative of a composite function, you usually apply the chain rule, which states that the derivative of a composite function can be found by:

• taking the derivative of the outer function
• multiplying it by the derivative of the inner function

In the problem, recognizing the composite nature was essential to simplify the task before differentiation. Once you detect the layers, like the function inside the logarithm and the exponential function itself, you are primed to use appropriate properties and rules, like those used for derivatives of exponential and logarithmic functions.

Logarithm Properties

Understanding logarithm properties can significantly simplify complicated expressions like in our example, where we have \(y = \ln(3\theta e^{-\theta})\). One of the most important properties of logarithms utilized here is that of the product rule: \( \ln(ab) = \ln(a) + \ln(b) \).This rule lets us break down the complex logarithmic expression into simpler parts that are easier to handle. Thus, \( \ln(3\theta e^{-\theta}) \) can be rewritten as \( \ln(3\theta) + \ln(e^{-\theta}) \).
Another key property is the logarithm of an exponential, where the log of \(e\) can essentially "cancel out" the base \(e\), leaving just the exponent. In our case, \( \ln(e^{- heta}) = -\theta \).These properties help streamline the differentiation process, leaving you with simpler terms to differentiate.

Exponential Function

Exponential functions are characterized by their constant base raised to a variable exponent, commonly expressed as \(e^x\). They have unique properties, most notably that their rate of change is proportional to their current value. This property is what makes exponential functions such a fascinating topic in calculus.
When dealing with exponential functions in derivatives, particularly where the base is \(e\), you benefit from one of the simplest derivative rules: the derivative of \(e^x\) is also \(e^x\). In the given exercise, we encounter \(e^{-\theta}\). Even though it's part of a composite function, breaking it down with logarithms leads us to simply differentiate with respect to \(\theta\).
This simplification to \(-\theta\) after applying the logarithm properties aids in differentiating, making it easier to grasp and execute.