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Math Help - Finding the value of the derivative.

  1. #1
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    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?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jeavus View Post
    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?
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  3. #3
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    I think I'm having problems simplifying the derivative equation with the sqroot(x+1)...
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  4. #4
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    What is the derivative that you have so far?
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  5. #5
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    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?
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  6. #6
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    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.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jeavus View Post
    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?
    Last edited by Jhevon; February 10th 2008 at 10:28 AM.
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