Answer :
[tex]f'(x_0)=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{8}{x};\ x_0=-1\\\\subtitute\\\\f'(-1)=\lim\limits_{x\to-1}\dfrac{\frac{8}{x}-\frac{8}{-1}}{x-(-1)}=\lim\limits_{x\to -1}\dfrac{\frac{8}{x}+8}{x+1}=\lim\limits_{x\to-1}\dfrac{\frac{8}{x}+\frac{8x}{x}}{x+1}\\\\=\lim\limits_{x\to-1}\dfrac{\frac{8+8x}{x}}{x+1}=\lim\limits_{x\to-1}\left(\dfrac{8+8x}{x}\cdot\dfrac{1}{x+1}\right)\\\\=\lim\limits_{x\to-1}\dfrac{8(x+1)}{x(x+1)}=\lim\limits_{x\to-1}\dfrac{8}{x}=\dfrac{8}{-1}=-8[/tex]