Problem 7 - Single Variable Calculus

Question

Differentiate the function. $$ y=x(\ln x)^{2} $$

Step-by-Step Solution

Answer

The derivative of the function \(y=x(\ln x)^{2}\) is: \(y' = (\ln x)^{2} + 2 (\ln x)\).

Step 1: Identify the functions u and v

Let the first function be \(u = x\) and the second function be \(v = (\ln x)^{2}\).

Step 2: Differentiate u and v separately

We will now differentiate each function. Differentiate \(u = x\): \(u' = \frac{d}{dx}x = 1\) Differentiate \(v = (\ln x)^{2}\): We can rewrite it as \(v = (g \circ h)(x)\), where \(g(x) = x^{2}\) and \(h(x) = \ln x\). Now, we will differentiate using the chain rule \(\frac{dv}{dx} = \frac{dg}{dh} \cdot \frac{dh}{dx}\). Differentiate \(g(x) = x^{2}\) with respect to \(h\): \(\frac{dg}{dh} = 2h\) Differentiate \(h(x) = \ln x\): \(\frac{dh}{dx} = \frac{1}{x}\) Now, using the chain rule: \(v' = \frac{dv}{dx} = \frac{dg}{dh} \cdot \frac{dh}{dx} = 2h\cdot\frac{1}{x}\), where \(h = \ln x\) So, \(v' = 2 (\ln x) \cdot \frac{1}{x}\)

Step 3: Apply the product rule

Now, we will apply the product rule using our information in \(u\) and \(v\) and their derivatives. \[(uv)' = u'v + uv' = 1\cdot(\ln x)^{2} + x\cdot 2 (\ln x) \cdot \frac{1}{x}\]

Step 4: Simplify the expression

Simplifying the expression, we get: \[\frac{d}{dx}(x(\ln x)^{2}) = (\ln x)^{2} + 2 (\ln x)\] Thus, the derivative of the function \(y=x(\ln x)^{2}\) is: \[y' = (\ln x)^{2} + 2 (\ln x)\]

Key Concepts

Understanding the Product Rule

The product rule is a key concept in differentiation, especially when dealing with functions that are products of two or more separate functions. Basically, it allows us to differentiate products of functions efficiently. Given two functions, say \( u(x) \) and \( v(x) \), their product \( y(x) = u(x) \cdot v(x) \) can be differentiated using the product rule. The formula to remember is:

• \( (uv)' = u'v + uv' \)

This means, to find the derivative of the product of two functions, you take:- The derivative of the first function \( u' \) multiplied by the second function \( v \),- Plus the first function \( u \) multiplied by the derivative of the second function \( v' \).
Applying this in our example where \( u(x) = x \) and \( v(x) = (\ln x)^2 \):

• Differentiate \( u \) to get \( u' = 1 \).
• Use the chain rule to differentiate \( v \) and get \( v' = \frac{2(\ln x)}{x} \).

Finally, plug these into the product rule formula: \( (uv)' = 1 \cdot (\ln x)^2 + x \cdot \frac{2(\ln x)}{x} \) resulting in \( (\ln x)^2 + 2(\ln x) \).

Learning the Chain Rule

The chain rule is an essential differentiation tool used when dealing with composite functions - functions within functions. This is the technique we use when we need to find the derivative of a function that can be expressed as \( g(h(x)) \). The chain rule states that:

• \( \frac{d}{dx}[g(h(x))] = g'(h(x)) \cdot h'(x) \)

In the exercise, \( v(x) = (\ln x)^2 \) is a composite function where \( g(h) = h^2 \) and \( h(x) = \ln x \).
To apply the chain rule:- First, differentiate the outer function \( g(h) = h^2 \) with respect to \( h \), yielding \( g'(h) = 2h \).- Then, differentiate the inner function \( h(x) = \ln x \) with respect to \( x \), giving us \( h'(x) = \frac{1}{x} \).
Altogether, according to the chain rule, the derivative of \( v(x) \) is:

• \( v' = 2 \ln x \cdot \frac{1}{x} \)

Derivative: The Core Idea

Differentiation is a fundamental concept in calculus, which deals with how a function changes as its input changes. When you differentiate a function, you are looking for its derivative—a new function that gives you the rate at which the original function changes.
Derivatives can be thought of in several ways, such as:

• Measures of how steep a graph is at any point (the slope)
• Describes how a system or function changes regarding its inputs

In our exercise, we differentiated the function \( y = x(\ln x)^2 \), and found its derivative: \( y' = (\ln x)^2 + 2\ln x \). This tells us how quickly or slowly \( y \) changes as \( x \) changes, providing insights into its rate of change at any point along the curve.

Single Variable Calculus Simplified

Single variable calculus focuses on functions that depend on one variable. It's a branch of calculus where differentiation and integration are applied to functions of one variable, such as \( y = f(x) \).
Key components of single variable calculus include:

• : Deals with finding the derivative, which describes how functions change.
• : Concerns with integration, the reverse process of differentiation.

Within this topic, the main goal is often to find rates of change or to understand how a quantity accumulates over time. Our example covered differentiation, where we applied both the product rule and chain rule to find the derivative of a single-variable function \( y \). Understanding these concepts is fundamental to mastering single-variable calculus.