1. **State the problem:** Find the derivative with respect to $x$ of the function $$f(x) = \sqrt{2x^2 + 3x - 4}.$$\n\n2. **Recall the formula:** The derivative of $\sqrt{u}$ with respect to $x$ is given by $$\frac{d}{dx} \sqrt{u} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}.$$\nHere, $u = 2x^2 + 3x - 4$.\n\n3. **Find $\frac{du}{dx}$:** Differentiate $u$ term-by-term:\n$$\frac{du}{dx} = \frac{d}{dx}(2x^2) + \frac{d}{dx}(3x) - \frac{d}{dx}(4) = 4x + 3 - 0 = 4x + 3.$$\n\n4. **Apply the chain rule:**\n$$\frac{d}{dx} \sqrt{2x^2 + 3x - 4} = \frac{1}{2\sqrt{2x^2 + 3x - 4}} \cdot (4x + 3).$$\n\n5. **Final answer:**\n$$\boxed{\frac{4x + 3}{2\sqrt{2x^2 + 3x - 4}}}.$$\n\nThis matches option 2.