Answer:
By definition, the derivative of f(x) is
[tex]lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]
Let's use the definition for [tex]f(x)=\frac{1}{x}[/tex]
[tex]lim_{h\rightarrow 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\\lim_{h\rightarrow 0} \frac{\frac{x-(x+h)}{x(x+h)}}{h}=\\lim_{h\rightarrow 0} \frac{\frac{(-1)h}{x^2+xh}}{h}=\\lim_{h\rightarrow 0} \frac{(-1)h}{h(x^2+xh)}=\\lim_{h\rightarrow 0} \frac{-1}{x^2+xh)}=\frac{-1}{x^2+x*0}=\frac{-1}{x^2}[/tex]
Then, [tex]f'(x)=\frac{-1}{x^2}[/tex]
