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?
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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?
I think I'm having problems simplifying the derivative equation with the sqroot(x+1)...
What is the derivative that you have so far?
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?
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: EDIT: Added lim so it was correct.
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 and remember, you are taking the limit as EDIT: hehe, we seem to be tripping ovr each other to answer the question, eh, Trevor?
Last edited by Jhevon; February 10th 2008 at 10:28 AM.
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