Answer
$ f'(8)=-\frac{1}{64}$
Equation of the tangent line:
$ y=-\frac{1}{64}x+\frac{1}{4}$
Work Step by Step
Recall that $ f'(a)=\lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$
Given that $ a=8$. Therefore, $ f'(8)=\lim\limits_{h \to 0}\frac{f(8+h)-f(8)}{h}=\lim\limits_{h \to 0}\frac{(8+h)^{-1}-(8^{-1})}{h}$
$=\lim\limits_{h \to 0}\frac{\frac{-h}{8(8+h)}}{h}=\lim\limits_{h \to 0}\frac{-1}{8(8+h)}=-\frac{1}{64}$
Equation of the tangent line is of the form
$ y-f(a)=f'(a)(x-a)$
Knowing that $ f(a)=f(8)=\frac{1}{8}, f'(a)=f'(8)=-\frac{1}{64}$ and $ a=8$, we obtain the equation of the tangent line through $ a $ as below.
$ y-\frac{1}{8}=-\frac{1}{64}(x-8)$
Or $ y=-\frac{1}{64}x+\frac{1}{4}$
