We are given the function:

$y = x^2 \ln x^2$

Step 1: Differentiate using the product rule

Since this function is a product of two functions, we use the product rule, which states:

$(uv)' = u'v + uv'$

Let:

• $u = x^2$, so $u' = 2x$
• $v = \ln x^2$, so we differentiate it separately.

Step 2: Differentiate $v = \ln x^2$

Using the logarithm property:

$\ln x^2 = 2 \ln x$

Thus,

$\frac{d}{dx} (\ln x^2) = 2 \cdot \frac{d}{dx} (\ln x) = \frac{2}{x}$

Step 3: Apply the product rule

Now, applying the product rule:

$y' = (2x)(\ln x^2) + (x^2) \left( \frac{2}{x} \right)$

$y' = 2x \ln x^2 + \frac{2x^2}{x}$

$y' = 2x \ln x^2 + 2x$

Final Answer:

$y' = 2x ( \ln x^2 + 1 )$

Would you like a step-by-step breakdown or another example?

Here are five related questions:

1. How do you differentiate $y = x^3 \ln x^3$?
2. What is the derivative of $y = x^n \ln x$ for any exponent $n$?
3. How does the chain rule apply when differentiating logarithmic functions?
4. What are some common mistakes when differentiating functions involving logarithms?
5. How do we handle differentiation when there are multiple logarithms in an expression?

Tip: Always check if logarithmic properties can simplify the function before differentiating!