To find the inflection point of the function \( y = x^2 \ln(x) \), we need to follow these steps:

### Step 1: Find the first derivative of \( y \)
We'll apply the product rule to differentiate the function.

The product rule states:
\[
\frac{d}{dx} (u \cdot v) = u'v + uv'
\]
Let:
- \( u = x^2 \) (with derivative \( u' = 2x \))
- \( v = \ln(x) \) (with derivative \( v' = \frac{1}{x} \))

Now apply the product rule:
\[
y' = \frac{d}{dx} \left( x^2 \ln(x) \right) = (2x)(\ln(x)) + (x^2)\left(\frac{1}{x}\right)
\]
Simplifying:
\[
y' = 2x \ln(x) + x
\]

### Step 2: Find the second derivative of \( y \)
Differentiate the first derivative \( y' = 2x \ln(x) + x \).

Again, apply the product rule to \( 2x \ln(x) \):

\[
\frac{d}{dx}(2x \ln(x)) = (2)(\ln(x)) + (2x)\left(\frac{1}{x}\right)
= 2 \ln(x) + 2
\]

The derivative of \( x \) is simply 1, so the second derivative becomes:
\[
y'' = 2 \ln(x) + 2 + 1
\]
Simplifying:
\[
y'' = 2 \ln(x) + 3
\]

### Step 3: Set the second derivative equal to zero
To find the inflection points, set the second derivative equal to zero and solve for \( x \):
\[
2 \ln(x) + 3 = 0
\]
\[
2 \ln(x) = -3
\]
\[
\ln(x) = -\frac{3}{2}
\]
Exponentiate both sides to solve for \( x \):
\[
x = e^{-\frac{3}{2}}
\]
Thus, the inflection point occurs at \( x = e^{-\frac{3}{2}} \).

### Conclusion:
The function \( y = x^2 \ln(x) \) has an inflection point at \( x = e^{-\frac{3}{2}} \).

Would you like more details or further clarification?

Here are 5 related questions for further practice:
1. Find the first derivative of \( y = x^2 \ln(2x) \).
2. Determine the inflection point of \( y = x^3 \ln(x) \).
3. What is the behavior of \( y = x^2 \ln(x) \) as \( x \to 0^+ \)?
4. Compute the concavity of \( y = x^2 \ln(x) \) at \( x = 1 \).
5. Find the critical points of \( y = x^2 \ln(x) \).

**Tip:** The product rule is essential when differentiating functions involving two multiplicative terms like \( x^2 \ln(x) \).

Learn how to find the inflection point of the function y = x^2 ln(x) through step-by-step differentiation, applying the product rule, and setting the second derivative to zero. This problem is suitable for advanced high school students or first-year calculus learners.

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