How do you find the definition of f(x)=x (sq. root(x+1))
How far did you get?
$\displaystyle f'(x)=\lim_{h\to 0}\frac{(x+h)\sqrt{x+h+1}-x\sqrt{x+1}}{h}$
I'm sure you got stuck here. The thing we need to do now is multiply by the conjugate of the numerator:
$\displaystyle f'(x)=\lim_{h\to 0}\frac{(x+h)\sqrt{x+h+1}-x\sqrt{x+1}}{h}\cdot\frac{(x+h)\sqrt{x+h+1}+x\sqrt {x+1}}{(x+h)\sqrt{x+h+1}+x\sqrt{x+1}}$
This simplifies to $\displaystyle \lim_{h\to 0}\frac{(x+h)^2(x+h+1)-x^2(x+1)}{h\left[(x+h)\sqrt{x+h+1}+x\sqrt{x+1}\right]}$
Expand out the numerator, and try to cancel out as many terms as you can. Then a common factor of h should appear in all the numerator terms. Pull it out, and it should then cancel out with the h term in the denominator. Then evaluate the limit.
Can you try to take it from here?
--Chris