Answer
$f'(2)=-\dfrac{1}{9}$
Work Step by Step
$f(x)=\dfrac{1}{x+1},$ at $2$
The derivative of a function $f$ at $a$ is $f'(a)=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$
In this case, $a=2$
Find $f(a+h)$ by substituting $x$ by $2+h$ in $f(x)$ and simplifying:
$f(2+h)=\dfrac{1}{2+h+1}=\dfrac{1}{3+h}$
Find $f(a)$ by substituting $x$ by $2$ in $f(x)$ and evaluating:
$f(2)=\dfrac{1}{2+1}=\dfrac{1}{3}$
Substitute the known values into the formula that gives $f'(a)$ and evaluate:
$f'(a)=\lim_{h\to0}\dfrac{\dfrac{1}{3+h}-\dfrac{1}{3}}{h}=\lim_{h\to0}\dfrac{\dfrac{3-(3+h)}{3(3+h)}}{h}=...$
$...=\lim_{h\to0}\dfrac{3-3-h}{3h(3+h)}=\lim_{h\to0}\dfrac{-h}{3h(3+h)}=\lim_{h\to0}\dfrac{-1}{3(3+h)}=...$
$...=\dfrac{-1}{3(3+0)}=-\dfrac{1}{9}$
