We are asked to differentiate the function:

$f(x) = e^{(3x^2 - 2x + 4)^4}$

To differentiate this function, we will need to apply both the chain rule and the product of derivatives (since we have an exponential function with a composite argument).

Step 1: Let $u = (3x^2 - 2x + 4)^4$, so that $f(x) = e^u$.

We now need to differentiate $f(x) = e^u$ with respect to $x$, applying the chain rule:

$\frac{d}{dx}\left( e^u \right) = e^u \cdot \frac{du}{dx}$

Step 2: Differentiate $u = (3x^2 - 2x + 4)^4$.

Here, we need to apply the chain rule again because $u$ is a composite function of $x$. Let $v = 3x^2 - 2x + 4$, so that $u = v^4$. The derivative of $u$ is:

$\frac{du}{dx} = \frac{d}{dx}(v^4) = 4v^3 \cdot \frac{dv}{dx}$

Now differentiate $v = 3x^2 - 2x + 4$:

$\frac{dv}{dx} = 6x - 2$

So, the derivative of $u$ becomes:

$\frac{du}{dx} = 4(3x^2 - 2x + 4)^3 \cdot (6x - 2)$

Step 3: Substitute back into the expression for $f'(x)$.

Now substitute $\frac{du}{dx}$ into the expression for $f'(x)$:

$f'(x) = e^{(3x^2 - 2x + 4)^4} \cdot 4(3x^2 - 2x + 4)^3 \cdot (6x - 2)$

Thus, the derivative of $f(x)$ is:

$f'(x) = 4e^{(3x^2 - 2x + 4)^4} (3x^2 - 2x + 4)^3 (6x - 2)$

---

Would you like further details on this differentiation process or have any other questions? Here are some related problems you might find helpful:

1. Differentiate $f(x) = e^{(x^3 - 5x + 2)^2}$.
2. Differentiate $f(x) = \ln((4x^2 + 3)^5)$.
3. Find the second derivative of $f(x) = e^{x^2 + 1}$.
4. Differentiate $f(x) = (2x^3 - 5x + 7)^3$.
5. Solve $f(x) = e^{(2x - 1)}$ and differentiate it.

Tip: Always remember to use the chain rule when differentiating composite functions!