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.