Answer
$f'(-3)=\dfrac{2}{25}$
Work Step by Step
$f(x)=\dfrac{x}{2-x},$ at $-3$
The derivative of a function $f$ at $a$ is $f'(a)=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$
In this case, $a=-3$
Find $f(a+h)$ by substituting $x$ by $-3+h$ in $f(x)$ and simplifying:
$f(-3+h)=\dfrac{-3+h}{2-(-3+h)}=\dfrac{-3+h}{2+3-h}=\dfrac{-3+h}{5-h}$
Find $f(a)$ by substituting $x$ by $-3$ in $f(x)$ and evaluating:
$f(-3)=\dfrac{-3}{2-(-3)}=\dfrac{-3}{2+3}=-\dfrac{3}{5}$
Substitute the known values into the formula that gives $f'(a)$ and evaluate:
$f'(-3)=\lim_{h\to0}\dfrac{\dfrac{-3+h}{5-h}+\dfrac{3}{5}}{h}=...$
$...=\lim_{h\to0}\dfrac{\dfrac{5(-3+h)+3(5-h)}{5(5-h)}}{h}=...$
$...=\lim_{h\to0}\dfrac{-15+5h+15-3h}{5h(5-h)}=\lim_{h\to0}\dfrac{2h}{5h(5-h)}=...$
$...=\lim_{h\to0}\dfrac{2}{5(5-h)}=\dfrac{2}{5(5-0)}=\dfrac{2}{25}$
