How do I find the derivative of #f(x) = sqrt(x+3)# using first principles?

1 Answer
Apr 20, 2018

#f'(x)=1/(2sqrt(x+3))#

Explanation:

#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))#