Problem 19 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The derivative of \( y \) with respect to \( t \) is \( \frac{1}{t} - 1 \).

Step 1: Recognize the Structure of the Function

The function given is a natural logarithm: \( y = \ln(3t e^{-t}) \). Notice that this involves a product of two functions inside the logarithm: \( 3t \) and \( e^{-t} \). To differentiate, we will use both the chain rule and the properties of logarithms.

Step 2: Apply Logarithmic Properties

We can use the property of logarithms which states that \( \ln(ab) = \ln a + \ln b \). Here, let \( a = 3t \) and \( b = e^{-t} \). Thus, \( y = \ln(3t) + \ln(e^{-t}) \). Simplifying further gives: \( y = \ln(3t) - t \), because \( \ln(e^{-t}) = -t \).

Step 3: Differentiate Using Chain Rule and Basic Derivatives

Now, differentiate \( y = \ln(3t) - t \) with respect to \( t \). The derivative of \( \ln(3t) \) with respect to \( t \) is \( \frac{1}{3t} \times 3 = \frac{1}{t} \), since the derivative of \( \ln u \) is \( \frac{1}{u} \) with \( u = 3t \) and \( \frac{d}{dt}(3t) = 3 \). The derivative of \(-t\) with respect to \( t \) is \(-1\).

Step 4: Combine the Derivatives

Combine the partial derivatives to find \( \frac{dy}{dt} = \frac{1}{t} - 1 \). This is the derivative of \( y \) with respect to \( t \).

Key Concepts

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate compositions of functions. It helps when one function is nested inside another. For instance, in our problem, we have a natural logarithmic function involving a product of functions: \( \ln(3t e^{-t}) \).
To apply the chain rule, consider the function inside the logarithm as \( u = 3t e^{-t} \). The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). However, since \( u \) itself is a function of \( t \), we must multiply by \( \frac{du}{dt} \) to complete the differentiation.
To find \( \frac{du}{dt} \), apply the product rule to \( u = 3t e^{-t} \). We will delve deeper into the product rule next. Ultimately, the chain rule allows us to handle the compound structure by breaking the problem into simpler differentiable parts.

Product Rule

The product rule is another differentiation technique that deals with functions that are multiplied together. It's key when differentiating the product \( 3t e^{-t} \). This function is composed of two separate parts \( f(t) = 3t \) and \( g(t) = e^{-t} \).
The product rule states: if \( h(t) = f(t) \times g(t) \), then its derivative \( h'(t) \) is \( f'(t)g(t) + f(t)g'(t) \). In our case, find the derivatives \( f'(t) = 3 \) and \( g'(t) = -e^{-t} \) (because the derivative of \( e^{-t} \) is \(-e^{-t}\)).
Thus, applying the product rule gives: \( \frac{d}{dt}[3t e^{-t}] = 3e^{-t} - 3t e^{-t} \). This derivative synonymously aligns with the chain rule approach, helping solve the problem effectively.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithmic function with a base of \( e \), where \( e \approx 2.718 \). It has various properties that simplify differentiation, particularly \( \ln(ab) = \ln a + \ln b \) and \( \ln\left(\frac{a}{b}\right)=\ln a - \ln b \).
For differentiation purposes, the derivative of \( \ln x \) is \( \frac{1}{x} \). This is crucial when dealing with functions like \( y = \ln(3t) - t \).
Incorporating properties of natural logarithms can simplify complex functions into sums or differences of simpler logarithmic terms, as shown. This simplification permits easier use of derivative rules, avoiding the need for the chain rule had the properties not been applied.

Derivative Rules

Derivative rules encompass a set of guidelines for differentiating various types of functions. Familiar ones include the power rule, product rule, and chain rule, each applicable to specific situations.
For example:

• Power Rule: \( \frac{d}{dx}[x^n] = nx^{n-1} \)
• Constant Rule: \( \frac{d}{dx}[c] = 0 \)
• Sum Rule: \( \frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x) \)

In our exercise, after splitting the logarithmic function, \( y = \ln(3t) - t \), derivative rules are straightforwardly applied to each component.
The differentiation yields \( \frac{1}{t} \) from \( \ln(3t) \) due to the chain rule application and \(-1\) from \(-t\). Combining them, we're left with \( \frac{dy}{dt} = \frac{1}{t} - 1 \). Understanding these fundamental rules allows students to tackle complex problems more confidently.