To find the derivative of the function \( f(x) = x^2 \cdot \sqrt{1 - x^2} \), we'll use both the **product rule** and the **chain rule**. 

### Step-by-Step Solution:

1. **Identify the function components:**

   - Let \( u(x) = x^2 \)
   - Let \( v(x) = \sqrt{1 - x^2} \)

2. **Apply the product rule**:

   The product rule states that for functions \( u(x) \) and \( v(x) \):
   \[
   f'(x) = u'(x)v(x) + u(x)v'(x)
   \]

3. **Differentiate \( u(x) = x^2 \)**:

   \[
   u'(x) = 2x
   \]

4. **Differentiate \( v(x) = \sqrt{1 - x^2} \) using the chain rule**:

   - First, express \( v(x) \) as \( (1 - x^2)^{1/2} \).
   - The derivative of \( v(x) \) with respect to \( x \) is:
     \[
     v'(x) = \frac{1}{2} \cdot (1 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}}
     \]

5. **Substitute into the product rule**:

   \[
   f'(x) = u'(x)v(x) + u(x)v'(x)
   \]
   \[
   f'(x) = (2x) \cdot \sqrt{1 - x^2} + x^2 \cdot \left(\frac{-x}{\sqrt{1 - x^2}}\right)
   \]

6. **Simplify the expression**:

   \[
   f'(x) = 2x\sqrt{1 - x^2} - \frac{x^3}{\sqrt{1 - x^2}}
   \]

   You can leave the derivative in this form, but it can also be simplified further by combining like terms if needed.

### Final Answer:
\[
f'(x) = 2x\sqrt{1 - x^2} - \frac{x^3}{\sqrt{1 - x^2}}
\]

Would you like more details or have any questions?

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### Related Questions:
1. How do you simplify the derivative further if possible?
2. What are some examples of other functions where the product rule is applied?
3. Can you explain the chain rule in more depth with a different function?
4. What happens to the derivative \( f'(x) \) at \( x = 1 \) or \( x = -1 \)?
5. How does the graph of the derivative function compare to the original function?

### Tip:
When using the product rule, always make sure to clearly differentiate between the two functions and apply the chain rule correctly if one of the functions is composite.

Learn how to find the derivative of f(x) = x²∙√(1 - x²) using the product rule and chain rule. This detailed step-by-step solution covers differentiation techniques, including the product rule and chain rule, suitable for advanced high school students and beyond.

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