# Thread: Finding the value of the derivative.

1. ## Finding the value of the derivative.

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?

2. Originally Posted by Jeavus
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?
do you know how to find derivatives?

the question is asking for the derivative evaluated at the point x = 0

so you want to find f'(0)

can you continue?

3. I think I'm having problems simplifying the derivative equation with the sqroot(x+1)...

4. What is the derivative that you have so far?

5. f(x)=sqroot(x+1); a=0

f'(x)=(f(x+h)-f(x))/h
f'(0)=(sqroot(x+1+h)-sqroot(x+1))/h

How do I simplify that?

6. 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:

$f'(x) = \lim_{h \rightarrow 0} \frac{\sqrt{x + 1 + h} - \sqrt{x + 1}}{h} \left( \frac{\sqrt{x + 1 + h} + \sqrt{x + 1}}{\sqrt{x + 1 + h} + \sqrt{x + 1}} \right)$

EDIT: Added lim so it was correct.

7. Originally Posted by Jeavus
f(x)=sqroot(x+1); a=0

f'(x)=(f(x+h)-f(x))/h
f'(0)=(sqroot(x+1+h)-sqroot(x+1))/h

How do I simplify that?
oh, you have to do it by the limit definition. multiply by the conjugate over itself. that is, multiply by $\frac {\sqrt{x + 1 + h} + \sqrt{x + 1}}{\sqrt{x + 1 + h} + \sqrt {x + 1}}$

and remember, you are taking the limit as $h \to 0$

EDIT: hehe, we seem to be tripping ovr each other to answer the question, eh, Trevor?