Dùng định nghĩa để tính đạo hàm của các hàm số sau:

c)  $f\left(x\right)=\frac{1}{x+1}$;

c) Với  $x_0\ne -1$, ta có:

 $y^{\text{'}}\left(x_0\right)=\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}=\underset{x\to x_0}{\lim }\frac{\frac{1}{x+1}-\frac{1}{x_0+1}}{x-x_0}$

 $=\underset{x\to x_0}{\lim }\frac{\frac{1}{x+1}-\frac{1}{x_0+1}}{x-x_0}=\underset{x\to x_0}{\lim }\frac{\left(x_0+1\right)-\left(x+1\right)}{\left(x-x_0\right)\left(x+1\right)\left(x_0+1\right)}$

 $=\underset{x\to x_0}{\lim }\frac{x_0-x}{\left(x-x_0\right)\left(x+1\right)\left(x_0+1\right)}==\underset{x\to x_0}{\lim }-\frac{1}{\left(x+1\right)\left(x_0+1\right)}$

 $=-\frac{1}{{\left(x_0+1\right)}^2}$.

Vậy  $y^{\text{'}}\left(x\right)=-\frac{1}{{\left(x+1\right)}^2}\left(x\ne -1\right).$