Problem 11 - Applied Calculus

Question

Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ P=t^{2} \ln t $$

Step-by-Step Solution

Answer

The derivative of the function is \(P' = 2t \ln t + t\).

Step 1: Identify the Functions

The given function is the product of two functions: \( f(t) = t^2 \) and \( g(t) = \ln t \). To find the derivative, we will use the product rule for differentiation, which is used when differentiating products of two functions.

Step 2: Recall the Product Rule

The product rule states that if you have two functions \( f(t) \) and \( g(t) \), then the derivative of their product \( P = f(t) \, g(t) \) is given by: \[P' = f'(t) \, g(t) + f(t) \, g'(t)\]

Step 3: Differentiate \(f(t)\) and \(g(t)\)

Find the derivative of \(f(t) = t^2\):- \(f'(t) = 2t\).Next, find the derivative of \(g(t) = \ln t\):- \(g'(t) = \frac{1}{t}\).

Step 4: Apply the Product Rule

Using the product rule from Step 2, substitute \(f(t)\), \(f'(t)\), \(g(t)\), and \(g'(t)\):\[P' = (2t)(\ln t) + (t^2)\left(\frac{1}{t}\right)\]Simplify this expression to find:\[P' = 2t \ln t + t\]

Step 5: Simplify and Write the Final Answer

Combine like terms to simplify the expression further:- Since there's nothing more to combine or simplify, the derivative \(P'\) is\[P' = 2t \ln t + t\]

Key Concepts

Product Rule for Differentiation

The product rule is a fundamental tool in calculus when dealing with the derivative of the product of two functions.
This rule comes in handy when you need to find the derivative of a function like our example, where the function is a product of two separate functions. In our exercise, these functions were \(f(t) = t^2\) and \(g(t) = \ln t\).
In simple terms, the product rule states:

• Take the derivative of the first function and multiply it by the second function.
• Then take the derivative of the second function and multiply it by the first function.

Combine these two results to complete the differentiation. The mathematical expression of the product rule is:
\[P' = f'(t) \cdot g(t) + f(t) \cdot g'(t)\] where \(P\) is the product of \(f(t)\) and \(g(t)\). Using this method simplifies handling complex expressions by breaking them down into simpler, manageable parts. The importance of the product rule lies in its ability to expand the scope of differentiation beyond single-variable functions.

Derivative Calculation Process

To calculate a derivative, it is essential to understand and correctly apply differentiation rules, such as the product rule mentioned earlier. Calculating derivatives involves several steps:

• Identify the component functions that make up the product, as we did with \(f(t) = t^2\) and \(g(t) = \ln t\).
• Determine the derivative of each individual function one at a time:
• For \(f(t) = t^2\), the derivative \(f'(t)\) is identified using the power rule, resulting in \(2t\).
• For \(g(t) = \ln t\), the derivative \(g'(t)\) uses the rule for logarithmic functions, giving \(\frac{1}{t}\).
• Scrupulously apply the product rule to these derivatives to calculate the overall derivative of the function.
• Simplify the resulting expression to find the cleanest possible form of the derivative.

In our exercise, after calculating \(f'(t)\) and \(g'(t)\), substituting these into the product rule yielded \(P' = 2t \ln t + t\), a process that underscores the structured logic of calculus differentiation.

Differentiation of the Natural Logarithm

Differentiating expressions that involve the natural logarithm, \(\ln t\), calls for a specific rule. This rule states that the derivative of \(\ln t\) is \(\frac{1}{t}\). The natural logarithm function, being a fundamental mathematical function, has a unique, straightforward derivative which simplifies the differentiation process for expressions involving it.

Here's how it plays out:

• The derivative \(\frac{1}{t}\) is derived from recognizing the rate at which \(\ln t\) changes with respect to \(t\).
• This is essential because \(\ln t\) grows at a decreasing rate as \(t\) increases, influencing how we calculate other connected derivatives.
• Understanding this rule aids in tackling any calculus problems involving logarithmic differentiation, forming a key part of solving complex expressions with logarithmic components.

For our specific exercise, recognizing this rule enabled us to accurately apply it within the broader differentiation context using the product rule, leading to the successful calculation of the overall derivative. It underscores the importance of knowing individual differentiation rules and how they interconnect in calculus.