#f'(x)=lim_(h->0)(f(x+h)-f(x))/h#
#f(x)=sqrt(x+3), f(x+h)=sqrt(x+h+3)#, then
#f'(x)=lim_(h->0)(sqrt(x+h+3)-sqrt(x+3))/h#
If we evaluate this right away, we get
#lim_(h->0)(sqrt(x+h+3)-sqrt(x+3))/h=(sqrt(x+3)-sqrt(x+3))/0=0/0#,
so we need to simplify as this is an indeterminate form.
Multiply the entire limit by the numerator's conjugate, which is #(sqrt(x+h+3)+sqrt(x+3))/(sqrt(x+h+3)+sqrt(x+3))#. This is the same as multiplying by #1.#
#f'(x)=lim_(h->0)(sqrt(x+h+3)-sqrt(x+3))/h*(sqrt(x+h+3)+sqrt(x+3))/(sqrt(x+h+3)+sqrt(x+3))#
The numerator becomes
#sqrt(x+h+3)-sqrt(x+3) * [sqrt(x+h+3)+sqrt(x+3)]=x+h+3-(x+3)=x+h+3-x-3=h#
#f'(x)=lim_(h->0)(cancelx+h+cancel3-cancelx-cancel3)/(h(sqrt(x+h+3)+sqrt(x+3))#
#f'(x)=lim_(h->0)(cancelh)/(cancelh(sqrt(x+h+3)+sqrt(x+3))#
#f'(x)=lim_(h->0)1/(sqrt(x+h+3)+sqrt(x+3))#
#f'(x)=1/(sqrt(x+3)+sqrt(x+3))#
#f'(x)=1/(2sqrt(x+3))#
