I've learned the short cut to obtain a derivate (nX^n-1) but some questions on exams ask to find the answer using the definition

Lim __f(x+h)-f(x)__

h->0 h

now I am terribly lost how to solve this is say f(x) = sqrt (x+2)

what I would think to do is rewrite it was

__(x+2)^1/2 + h) - (x+2)^1/2__

h

this is where I go astray...

I am totally baffled on what the next step would be...if it was x^2 I would simply factor it out and then divide by h but the 1/2 is totally making my thought process come to a dead halt. Please tell me if my approach is wrong or what the next step would be. Thanks

\(\displaystyle f(x) = \sqrt{x + 2}\)

\(\displaystyle f(x + h) = \sqrt{x + h + 2}\).

\(\displaystyle \frac{f(x + h) - f(x)}{h} = \frac{\sqrt{x + h + 2} - \sqrt{x + 2}}{h}\)

\(\displaystyle = \left(\frac{\sqrt{x + h + 2} - \sqrt{x + 2}}{h}\right)\left( \frac{\sqrt{x + h + 2}+\sqrt{x + 2}}{\sqrt{x + h + 2} + \sqrt{x + 2}}\right)\)

\(\displaystyle = \frac{(x + h + 2) - (x + 2)}{h(\sqrt{x + h + 2} + \sqrt{x + 2})}\)

\(\displaystyle = \frac{h}{h(\sqrt{x + h + 2} + \sqrt{x + 2})}\)

\(\displaystyle = \frac{1}{\sqrt{x + h + 2} + \sqrt{x + 2}}\).

So \(\displaystyle f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\)

\(\displaystyle = \lim_{h \to 0}\frac{1}{\sqrt{x + h + 2} + \sqrt{x + 2}}\)

\(\displaystyle = \frac{1}{\sqrt{x + 2} + \sqrt{x + 2}}\)

\(\displaystyle = \frac{1}{2\sqrt{x + 2}}\).