I'm asked to "Find the value of the derivative f'(a) for the given value of a.
f(x) = sqroot(x+1); a=0
How do I solve this?
Ok, I guess you haven't done the chain rule yet. You can do that with definition of the derivative. You just need to do the following:
$\displaystyle f'(x) = \lim_{h \rightarrow 0} \frac{\sqrt{x + 1 + h} - \sqrt{x + 1}}{h} \left( \frac{\sqrt{x + 1 + h} + \sqrt{x + 1}}{\sqrt{x + 1 + h} + \sqrt{x + 1}} \right)$
EDIT: Added lim so it was correct.
oh, you have to do it by the limit definition. multiply by the conjugate over itself. that is, multiply by $\displaystyle \frac {\sqrt{x + 1 + h} + \sqrt{x + 1}}{\sqrt{x + 1 + h} + \sqrt {x + 1}}$
and remember, you are taking the limit as $\displaystyle h \to 0$
EDIT: hehe, we seem to be tripping ovr each other to answer the question, eh, Trevor?