Ta có:
f(x) = $\frac{3x}{x^{2} + 1}$
=> f'(x) = ($\frac{3x}{x^{2} + 1}$)' = $\frac{(3x)'(x^{2} + 1) - 3x(x^{2} + 1)'}{(x^{2} + 1)^{2}}$
= $\frac{3(x)'(x^{2} + 1) - 3x[(x^{2})' + (1)']}{(x^{2} + 1)^{2}}$
= $\frac{3.1(x^{2} + 1) - 3x(2x + 0)}{(x^{2} + 1)^{2}}$
= $\frac{3x^{2} + 3 - 6x^{2}}{(x^{2} + 1)^{2}}$
= $\frac{-3x^{2} + 3}{(x^{2} + 1)^{2}}$
= $\frac{-3(x^{2} - 1)}{(x^{2} + 1)^{2}}$
= $\frac{-3(x - 1)(x+1)}{(x^{2} + 1)^{2}}$
