Problem 21 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=t(\ln t)^{2}$$

Step-by-Step Solution

Answer

The derivative is \(y' = (\ln t)^2 + 2\ln t\).

Step 1: Identify the differentiation rule

The function given is in the form of a product: \(y = t(\ln t)^2\), which consists of two functions multiplied together. Therefore, to differentiate this function with respect to \(t\), we will use the product rule. The product rule states that if \(u(t) = f(t)g(t)\), then \(u'(t) = f'(t)g(t) + f(t)g'(t)\). Here, \(f(t) = t\) and \(g(t) = (\ln t)^2\).

Step 2: Differentiate the first function

Now we'll differentiate \(f(t) = t\) with respect to \(t\). The derivative of \(t\) is 1, so \(f'(t) = 1\).

Step 3: Differentiate the second function

The second function is \(g(t) = (\ln t)^2\). To differentiate \( (\ln t)^2 \), we will use the chain rule. Let \( u = \ln t \), so \( g(t) = u^2 \), and \(\frac{du}{dt} = \frac{1}{t}\). Differentiating \(u^2\) with respect to \(u\), we get \(2u\). Applying the chain rule, \(g'(t) = 2u \cdot \frac{du}{dt} = 2(\ln t) \cdot \frac{1}{t} = \frac{2\ln t}{t}\).

Step 4: Apply the product rule

Substitute \(f(t)\), \(f'(t)\), \(g(t)\), and \(g'(t)\) into the product rule formula: \(y' = f'(t)g(t) + f(t)g'(t)\). We have \(f'(t) = 1\), \(g(t) = (\ln t)^2\), \(f(t) = t\), and \(g'(t) = \frac{2\ln t}{t}\). Therefore, \(y' = (1)(\ln t)^2 + t\left(\frac{2\ln t}{t}\right) = (\ln t)^2 + 2\ln t\).

Step 5: Simplify the expression

Combine like terms if necessary. In this case, \((\ln t)^2 + 2\ln t\) is already simplified. Ensure you understand each transformation made in the differentiation process.

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus for finding the derivative of the product of two functions. When you have a function that is the product of two other functions, like in the case of our exercise where the function is \(y = t(\ln t)^2\), the product rule helps us differentiate efficiently.

The rule states: if you have two functions \(f(t)\) and \(g(t)\), their product's derivative \(u'(t)\) can be expressed as:

• \(u'(t) = f'(t)g(t) + f(t)g'(t)\)

This formula tells us that the derivative of the product is the derivative of the first function times the second function plus the first function times the derivative of the second function.

In our exercise:

• \(f(t) = t\) and \(g(t) = (\ln t)^2\)
• So, \(u'(t) = 1 \, (\ln t)^2 + t \, \frac{2 \ln t}{t}\)
• Here, \(1(\ln t)^2 + t \left(\frac{2 \ln t}{t}\right) = (\ln t)^2 + 2\ln t\)

By applying the product rule, we've successfully broken down the differentiation into manageable parts.

Chain Rule

The chain rule is another essential tool in the differentiation toolkit. It is used when you need to differentiate a composite function, meaning a function that is composed of other functions. In our exercise, we used the chain rule when differentiating \((\ln t)^2\).

To apply the chain rule, we first identify the outer function and the inner function. For the function \((\ln t)^2\):

• The outer function is \(u^2\) and the inner function is \(u = \ln t\)

The chain rule states that to find the derivative, you first differentiate the outer function with respect to the inner function, and then multiply it by the derivative of the inner function:

• \(g'(t) = \frac{d}{du}(u^2) \cdot \frac{du}{dt}\)
• We have \(\frac{d}{du}(u^2) = 2u\), and \(\frac{du}{dt} = \frac{1}{t}\)
• Thus, \(g'(t) = 2u \cdot \frac{1}{t} = \frac{2\ln t}{t}\)

The chain rule helps us navigate through layers of functions smoothly and find derivatives effectively.

Natural Logarithm

The natural logarithm, denoted as \(\ln t\), is an intrinsic part of calculus, especially when dealing with exponential growth or decay. In our exercise, we worked with \((\ln t)^2\), which involves a natural logarithm.

Understanding how to differentiate functions involving \(\ln t\) is crucial:

• The derivative of \(\ln t\) with respect to \(t\) is \(\frac{1}{t}\)

This basic rule is key when applying both the product and chain rules, as seen in our steps. The natural logarithm has unique properties that make differentiating related functions straightforward.

It is important to remember that the natural logarithm gives us the time needed for a quantity to grow to a certain level at a constant growth rate. By applying its properties, combined with differentiation rules, we can solve complex calculus problems like the one in our exercise efficiently.