Derivar la función 3/x^2+1 aplicando el concepto o definición de derivada.

$$\begin{align}&\frac{d}{dy}\left(\frac{3}{x^{2}+1}\right)=\lim_{h\rightarrow0}{\frac{f(x+h)-f(x)}{h}}=\lim_{h\rightarrow0}{\frac{\frac{3}{(x+h)^{2}+1}-\frac{3}{x^{2}+1}}{h}}=\lim_{h\rightarrow0}{\left(\frac{3}{(x+h)^{2}+1}-\frac{3}{x^{2}+1}\right)}\frac{1}{h}=(3)\lim_{h\rightarrow0}{\left(\frac{1}{(x+h)^{2}+1}-\frac{1}{x^{2}+1}\right)}\frac{1}{h}\\&\\&...=(3)\lim_{h\rightarrow0}{\left(\frac{(x^{2}+1)-[(x+h)^{2}+1]}{[(x+h)^{2}+1](x^{2}+1)}\right)}\frac{1}{h}=(3)\lim_{h\rightarrow0}{\left(\frac{(x^{2}+1)-[x^{2}+2xh+h^{2}+1]}{[(x+h)^{2}+1](x^{2}+1)}\right)}\frac{1}{h}=(3)\lim_{h\rightarrow0}{\left(\frac{-h(2x+h)}{[(x+h)^{2}+1](x^{2}+1)}\right)}\frac{1}{h}\\&\\&...=(3)\lim_{h\rightarrow0}{\left(\frac{-2x-h}{[(x+h)^{2}+1](x^{2}+1)}\right)}=(3)\frac{-2x-(0)}{[(x+(0))^{2}+1](x^{2}+1)}=-\frac{6x}{(x^{2}+1)^{2}}\end{align}$$

y es todo