$b)x_0=4$  

$f^{{}^{\prime }}\left(4\right)=\underset{x\to 4}{\lim }\frac{f\left(x\right)-f\left(4\right)}{x-4}=\underset{x\to 4}{\lim }\frac{\frac{1}{x}-\frac{1}{4}}{x-4}=\underset{x\to 4}{\lim }\frac{\frac{4-x}{4x}}{x-4}=\underset{x\to 4}{\lim }\frac{4-x}{4x}\cdot \frac{1}{x-4}=\underset{x\to 4}{\lim }\frac{-1}{4x}=\frac{-1}{16}=-\frac{1}{16}$ $f^{{}^{\prime }}\left(4\right)=\underset{x\to 4}{\lim }\frac{f\left(x\right)-f\left(4\right)}{x-4}=\underset{x\to 4}{\lim }\frac{\frac{1}{x}-\frac{1}{4}}{x-4}=\underset{x\to 4}{\lim }\frac{\frac{4-x}{4x}}{x-4}=\underset{x\to 4}{\lim }\frac{4-x}{4x}\cdot \frac{1}{x-4}=\underset{x\to 4}{\lim }\frac{-1}{4x}=\frac{-1}{16}=-\frac{1}{16}$  

 

$c)x_0=-1$  

$f^{{}^{\prime }}\left(-1\right)=\underset{x\to -1}{\lim }\frac{f\left(x\right)-f\left(-1\right)}{x-\left(-1\right)}=\underset{x\to -1}{\lim }\frac{\frac{1}{x}+1}{x+1}=\underset{x\to -1}{\lim }\frac{\frac{1+x}{x}}{x+1}=\underset{x\to -1}{\lim }\frac{x+1}{x}\cdot \frac{1}{x+1}=\underset{x\to -1}{\lim }\frac{1}{x}=\frac{1}{-1}=-1$