## Calculus: Early Transcendentals (2nd Edition)

$f'(x)=-1$
$f(x)=\dfrac{4-x^{2}}{x-2}$ 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: $f'(x)=-(x)'-(2)'=-1$