Problem 7 - Applied Calculus

Question

Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ y=x \ln x $$

Step-by-Step Solution

Answer

The derivative of the function is \(1 + \ln x\).

Step 1: Identify the Derivative Rule to Use

The function given is the product of two simpler functions: \(x\) and \(\ln x\). The appropriate method to find the derivative of a product of functions is the product rule.

Step 2: Recall the Product Rule

The product rule states that the derivative of a product of two functions, \(u(x)v(x)\), is given by \(u'(x)v(x) + u(x)v'(x)\). Here, set \(u = x\) and \(v = \ln x\).

Step 3: Differentiate the First Function

Find the derivative of the first function: \(u = x\). The derivative is \(\frac{d}{dx}x = 1\).

Step 4: Differentiate the Second Function

Find the derivative of the second function: \(v = \ln x\). The derivative is \(\frac{d}{dx} \ln x = \frac{1}{x}\).

Step 5: Apply the Product Rule

Using the product rule, substitute \(u' = 1\), \(v = \ln x\), \(u = x\), and \(v' = \frac{1}{x}\) into the product rule formula: \(u'v + uv' = (1)(\ln x) + (x)(\frac{1}{x})\).

Step 6: Simplify the Expression

Simplify the expression: \(\ln x + 1\). Hence, the derivative of \(y = x \ln x\) is \(1 + \ln x\).

Key Concepts

Product Rule

When you need to differentiate a function that is a product of two or more simpler functions, the product rule is your go-to technique. Think of the product rule as a way to break down a problem.
To apply it:

• Identify the two functions being multiplied together. Let's call them \(u(x)\) and \(v(x)\).
• The product rule formula is \(u'(x)v(x) + u(x)v'(x)\). This tells us how to find the derivative of the product of these functions.

The idea is to take the derivative of the first function while keeping the second function constant. Then, take the derivative of the second function while keeping the first function constant, and add these results together.
An example where you apply the product rule is when finding the derivative of \(y = x \ln x\). Here, \(u = x\) and \(v = \ln x\). Using the product rule allows you to handle each part separately, making the differentiation process easier and more structured.

Natural Logarithm

The natural logarithm function, \(\ln x\), is an important function in many areas of math and science.
It is the logarithm to the base \(e\), where \(e \approx 2.71828\), a fundamental mathematical constant. If you have \(\ln x\), it gives the power to which \(e\) must be raised to equal \(x\).Here's why it's so useful:

• It is used for continuous growth models, like calculating compound interest or population growth.
• The natural logarithm has a simple derivative: the derivative of \(\ln x\) with respect to \(x\) is \(\frac{1}{x}\).

This simplicity makes it a favorite when differentiating. In mathematical functions, when you see a natural logarithm, you immediately leverage this easy derivative rule. Like in the function \(y = x \ln x\), the natural logarithm's easy differentiation facilitates the application of the product rule.

Differentiation Techniques

Differentiation is a core concept in calculus and involves finding how a function changes as its input changes, which we call the derivative.
Numerous techniques help us find derivatives, and mastering these can make calculations easier:

• The product rule is used when differentiating products of functions as discussed.
• The chain rule helps with functions nested within each other, making it ideal for complicated compositions of functions.
• The quotient rule is another tool for dividing one function by another when differentiating.

Each technique focuses on breaking down complex expressions into manageable steps. Choosing the right tool depends on the structure of the function. By recognizing when and how to apply these methods, like using the product rule for \(y = x \ln x\), differentiation becomes a more straightforward process.