#### Answer

$a.$ The value of the derivative is $f'(5) = -\frac{1}{100}$.
$b.$ The equation is $y=-\frac{1}{100}x+\frac{3}{20}$

#### Work Step by Step

$a.$ By definition of a derivative at point $a=5$ we have
$$f'(5)=\lim_{h\to0}\frac{f(5+h)-f(5)}{h}=\lim_{h\to0}\frac{\frac{1}{(5+h)+5}-\frac{1}{5+5}}{h}=\lim_{h\to0}\frac{\frac{1}{10+h}-\frac{1}{10}}{h}=\lim_{h\to0}\frac{\frac{10-(10+h)}{10(10+h)}}{h}=\lim_{h\to0}\frac{-h}{10h(10+h)}=\lim_{h\to0}\frac{-1}{10(10+h)}=\frac{-1}{10(10+0)}=-\frac{1}{100}.$$
$b.$ We know that the equation of the tangent at point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=5$, $f(a)=1/(5+5)=1/10$ and $f'(a)=-1/100$ we get
$$y-\frac{1}{10}=-\frac{1}{100}(x-5)\Rightarrow y-\frac{1}{10}=-\frac{1}{100}x+\frac{1}{20}$$ which gives $$y=-\frac{1}{100}x+\frac{1}{20}+\frac{1}{10}=-\frac{1}{100}x+\frac{3}{20}.$$