Problem 17 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

\(y=x \ln x^{2}\)

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = 2\ln(x) + 2\)

Step 1: Understand the Problem

Given the function: \(y = x \ln(x^2)\) Find the derivative \(\frac{{dy}}{{dx}}\).

Step 2: Use Logarithm Properties

Recall the logarithm property: \ln(a^b) = b \ln(a). Apply this to the given function: \(y = x \ln((x^2)) = x \cdot 2 \ln(x) = 2x \ln(x)\)

Step 3: Apply Product Rule

The function \(y = 2x \ln(x)\) can be derived using the product rule for derivatives. The product rule states that if you have two functions u(x) and v(x), then: \(\frac{{d}}{{dx}}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\). Here, let \(u(x) = 2x\) and \(v(x) = \ln(x)\).

Step 4: Differentiate each part

Find the derivatives of each part: \(u(x) = 2x\) \(u'(x) = 2\) \(v(x) = \ln(x)\) \(v'(x) = \frac{1}{x}\)

Step 5: Combine using the product rule

Using the product rule: \(\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)\) Substitute the values in: \(\frac{dy}{dx} = (2) \ln(x) + (2x) \frac{1}{x}\) Simplify: \(\frac{dy}{dx} = 2\ln(x) + 2\)

Key Concepts

Logarithm Properties

Logarithm properties are essential tools in calculus, especially when dealing with logarithmic functions. One key property to remember is \(\text{ln}(a^b) = b \text{ln}(a)\). This property is particularly useful because it allows us to simplify expressions involving logarithms raised to a power. In the original exercise, we applied this property to the function \(y = x \text{ln}(x^2)\). By using the logarithm property, we can rewrite the function as \(y = x \text{ln}(x^2) = x \times 2 \text{ln}(x) = 2x \text{ln}(x)\). Simplifying the function in this manner makes it easier to apply differentiation rules in subsequent steps.

Product Rule

The product rule in calculus is a fundamental technique for finding the derivative of a product of two functions. Mathematically, if you have two functions \(u(x)\) and \(v(x)\), the product rule states that \(\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\).

In our exercise, the function \(y = 2x \text{ln}(x)\) is a product of \(u(x) = 2x\) and \(v(x) = \text{ln}(x)\). Applying the product rule involves taking the derivatives of both \(u(x)\) and \(v(x)\) and then combining them according to the rule. This method ensures that we can systematically obtain the derivative of more complex functions that are expressed as products.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate at which the function's value changes. In the context of our given function \(y = 2x \text{ln}(x)\), we need to find the derivatives of each part of the product.

First, let's find the derivative of \(u(x) = 2x\). Using basic differentiation rules, we get \(u'(x) = 2\).

Next, for \(v(x) = \text{ln}(x)\), the derivative is \(\frac{d}{dx}[\text{ln}(x)] = \frac{1}{x}\).

Now apply the product rule:
\(\frac{d}{dx}[2x \text{ln}(x)] = 2 \times \text{ln}(x) + 2x \times \frac{1}{x}\).

Simplify to get:
\(\frac{dy}{dx} = 2 \text{ln}(x) + 2\).

Hence, the derivative of the original function \(y = x \text{ln}(x^2)\) is \(\frac{dy}{dx} = 2 \text{ln}(x) + 2\).