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We are tasked with finding \( f'(g(x)) \) given that:

\[
\frac{d}{dx} \left[ f(x^2 - 3x + 2) \right] = x^2 - 3x + 2
\]
and 
\[
g(x) = x^2 - 2x.
\]

### Solution Steps

1. **Differentiate \( f(x^2 - 3x + 2) \):**
   Since \( \frac{d}{dx} \left[ f(x^2 - 3x + 2) \right] = x^2 - 3x + 2 \), we apply the chain rule:
   \[
   \frac{d}{dx} f(u) = f'(u) \cdot \frac{du}{dx},
   \]
   where \( u = x^2 - 3x + 2 \). Differentiating \( u \) with respect to \( x \):
   \[
   \frac{d}{dx}(x^2 - 3x + 2) = 2x - 3.
   \]
   Thus, using the chain rule:
   \[
   f'(x^2 - 3x + 2) \cdot (2x - 3) = x^2 - 3x + 2.
   \]

2. **Solve for \( f'(x^2 - 3x + 2) \):**
   Dividing both sides by \( 2x - 3 \) (assuming \( 2x - 3 \neq 0 \)):
   \[
   f'(x^2 - 3x + 2) = \frac{x^2 - 3x + 2}{2x - 3}.
   \]

3. **Substitute \( g(x) = x^2 - 2x \):**
   We are asked to find \( f'(g(x)) \), and since \( g(x) = x^2 - 2x \), we substitute this into the expression for \( f' \):
   \[
   f'(g(x)) = \frac{(x^2 - 2x) - 3x + 2}{2x - 3}.
   \]
   Simplifying the numerator:
   \[
   f'(g(x)) = \frac{x^2 - 2x - 3x + 2}{2x - 3} = \frac{x^2 - 5x + 2}{2x - 3}.
   \]

Thus, the derivative \( f'(g(x)) \) is:

\[
f'(g(x)) = \frac{x^2 - 5x + 2}{2x - 3}.
\]

---

Would you like further clarifications or details? Here are some related questions to explore:

1. What is the significance of the chain rule in calculus?
2. Can we simplify the expression \( f'(g(x)) \) further?
3. What happens if \( 2x - 3 = 0 \) in the denominator of \( f'(g(x)) \)?
4. How do we apply the chain rule for composite functions in general?
5. Can we extend this method to higher-order derivatives?

**Tip:** Always be cautious of points where the denominator of a function could equal zero, as this may indicate singularities or undefined behavior.

Find f'(g(x)) if d/dx[f(x^2 - 3x + 2)] = x^2 - 3x + 2 and g(x) = x^2 - 2x.

This problem demonstrates how to find the derivative of a composite function f(g(x)) using the chain rule. Given the equation d/dx[f(x^2 - 3x + 2)] = x^2 - 3x + 2, and g(x) = x^2 - 2x, the solution walks through the steps of differentiation.

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