Given f(x) = (2x3)/(x4).

The derivative $f'(x)$ is defined as:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Substitute $f(x) = \frac{2}{x}$ and $f(x+h) = \frac{2}{x+h}$ into the formula.

$$f'(x) = \lim_{h \to 0} \frac{\frac{2}{x+h} - \frac{2}{x}}{h}$$

$$f'(x) = \lim_{h \to 0} \frac{\frac{2x - 2(x+h)}{x(x+h)}}{h}$$ $$f'(x) = \lim_{h \to 0} \frac{\frac{2x - 2x - 2h}{x(x+h)}}{h}$$ $$f'(x) = \lim_{h \to 0} \frac{\frac{-2h}{x(x+h)}}{h}$$

$$f'(x) = \lim_{h \to 0} \frac{-2h}{h \cdot x(x+h)}$$

Cancel out $h$ from the numerator and denominator.

$$f'(x) = \lim_{h \to 0} \frac{-2}{x(x+h)}$$

Now, substitute $h=0$ into the expression.

$$f'(x) = \frac{-2}{x(x+0)}$$ $$f'(x) = \frac{-2}{x^2}$$

The derivative of $f(x)$ is:

$$f'(x) = -\frac{2}{x^2}$$