Given f(x) = (7)/(x² + 3).

The derivative $f'(x)$ of a function $f(x)$ is given by:

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

Given $f(x) = \frac{7}{x^2 + 3}$.
Substitute $x+h$ into $f(x)$ to find $f(x+h)$:

$$f(x+h) = \frac{7}{(x+h)^2 + 3} = \frac{7}{x^2 + 2xh + h^2 + 3}$$

$$f(x+h) - f(x) = \frac{7}{x^2 + 2xh + h^2 + 3} - \frac{7}{x^2 + 3}$$

To subtract the fractions, find a common denominator:

$$= \frac{7(x^2 + 3) - 7(x^2 + 2xh + h^2 + 3)}{(x^2 + 2xh + h^2 + 3)(x^2 + 3)}$$

Expand the numerator:

$$= \frac{7x^2 + 21 - (7x^2 + 14xh + 7h^2 + 21)}{(x^2 + 2xh + h^2 + 3)(x^2 + 3)}$$ $$= \frac{7x^2 + 21 - 7x^2 - 14xh - 7h^2 - 21}{(x^2 + 2xh + h^2 + 3)(x^2 + 3)}$$

Simplify the numerator:

$$= \frac{-14xh - 7h^2}{(x^2 + 2xh + h^2 + 3)(x^2 + 3)}$$

Factor out $h$ from the numerator:

$$= \frac{h(-14x - 7h)}{(x^2 + 2xh + h^2 + 3)(x^2 + 3)}$$

$$\frac{f(x+h) - f(x)}{h} = \frac{\frac{h(-14x - 7h)}{(x^2 + 2xh + h^2 + 3)(x^2 + 3)}}{h}$$

Cancel out $h$:

$$= \frac{-14x - 7h}{(x^2 + 2xh + h^2 + 3)(x^2 + 3)}$$

$$f'(x) = \lim_{h \to 0} \frac{-14x - 7h}{(x^2 + 2xh + h^2 + 3)(x^2 + 3)}$$

Substitute $h=0$ into the expression:

$$f'(x) = \frac{-14x - 7(0)}{(x^2 + 2x(0) + (0)^2 + 3)(x^2 + 3)}$$ $$f'(x) = \frac{-14x}{(x^2 + 3)(x^2 + 3)}$$ $$f'(x) = \frac{-14x}{(x^2 + 3)^2}$$

The derivative of $f(x) = \frac{7}{x^2 + 3}$ from first principles is:

$$\frac{-14x}{(x^2 + 3)^2}$$