Step 1: Write down the definition of the derivative from first principles. The derivative $f'(x)$ of a function $f(x)$ is given by: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ Step 2: Identify $f(x)$ and $f(x+h)$. Given $f(x) = \frac{7}{x^2 + 3}$. Substitute $x+h$ into $f(x)$ to find $f(x+h)$: $$f(x+h) = \frac{7}{(x+h)^2 + 3} = \frac{7}{x^2 + 2xh + h^2 + 3}$$ Step 3: Calculate $f(x+h) - f(x)$. $$f(x+h) - f(x) = \frac{7}{x^2 + 2xh + h^2 + 3} - \frac{7}{x^2 + 3}$$ 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: $$= \frac{h(-14x - 7h)}{(x^2 + 2xh + h^2 + 3)(x^2 + 3)}$$ Step 4: Divide by $h$. $$\frac{f(x+h) - f(x)}{h} = \frac{\frac{h(-14x - 7h)}{(x^2 + 2xh + h^2 + 3)(x^2 + 3)}}{h}$$ Cancel out $h$: $$= \frac{-14x - 7h}{(x^2 + 2xh + h^2 + 3)(x^2 + 3)}$$ Step 5: Take the limit as $h \to 0$. $$f'(x) = \lim_{h \to 0} \frac{-14x - 7h}{(x^2 + 2xh + h^2 + 3)(x^2 + 3)}$$ 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: $$\boxed{\frac{-14x}{(x^2 + 3)^2}}$$