Calc2.57

From Exampleproblems

$f(x)=\frac{1}{x}\,$

$f'(x)=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}=\lim_{\Delta x \to 0}\frac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x}=\lim_{\Delta x \to 0}\frac{\frac{x-(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}=\lim_{\Delta x \to 0}\frac{\frac{-\Delta x}{x(x+\Delta x)}}{\Delta x}$

$=\lim_{\Delta x \to 0}\frac{-\Delta x}{x\Delta x(x+\Delta x)}=\lim_{\Delta x \to 0}\frac{-1}{x(x+\Delta x)}=-\frac{1}{x^2}$

Thus, we have shown that the derivative of $\frac{1}{x}$ is $-1\frac{1}{x^2}$ .