Answer
$f'(x)=-1$
Work Step by Step
$f(x)=\dfrac{4-x^{2}}{x-2}$
Factor the numerator, which is a difference of squares:
$f(x)=\dfrac{4-x^{2}}{x-2}=\dfrac{(2-x)(2+x)}{x-2}=...$
Change the sign of the numerator and the sign of the fraction and simplify:
$...=-\dfrac{(x-2)(x+2)}{x-2}=-(x+2)=-x-2$
Evaluate the derivative term by term:
$f'(x)=-(x)'-(2)'=-1$
