Let us use the definition:
#f'(x)=lim_(h->0)[f(x+h)-f(x)]/h#
where #h# is a small increment.
In our case we get:
#lim_(h->0)[sqrt(2-x-h)+1/(x+h-3)-sqrt(2-x)-1/(x-3)]/h#
let us use a rule of limits to write:
#lim_(h->0)[sqrt(2-x-h)-sqrt(2-x)]/h+lim_(h->0)[1/(x+h-3)-1/(x-3)]/h#
we can now operate on the first:
#lim_(h->0)[sqrt(2-x-h)-sqrt(2-x)]/h*[sqrt(2-x-h)+sqrt(2-x)]/[sqrt(2-x-h)+sqrt(2-x)]#
#lim_(h->0)[cancel(2)cancel(-x)-hcancel(-2)+cancel(x)]/h*1/[sqrt(2-x-h)+sqrt(2-x)]#
#lim_(h->0)-h/h*1/[sqrt(2-x-h)+sqrt(2-x)]#
#lim_(h->0)-1/[sqrt(2-x-h)+sqrt(2-x)]#
as #h->0# we get:
#lim_(h->0)-1/[sqrt(2-x-h)+sqrt(2-x)]=-1/[2sqrt(2-x)]#
Then we operate on the second:
#lim_(h->0)[1/(x+h-3)-1/(x-3)]/h=lim_(h->0)1/h[(cancel(x)cancel(-3)cancel(-x)-hcancel(+3))/((x+h-3)(x-3))]#
#lim_(h->0)-h/h[1/((x+h-3)(x-3))]#
#lim_(h->0)[-1/((x+h-3)(x-3))]#
as #h->0#
#lim_(h->0)[-1/((x+h-3)(x-3))]=-1/(x-3)^2#
and in conclusion:
#f'(x)=-1/[2sqrt(2-x)]-1/(x-3)^2#
