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Math Help - derivative of a function problem

  1. #1
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    derivative of a function problem

    i'm having a major brain clot,

    the derivative of this function

    f(x)= √x + 1




    using the limit defintion




    thank you so much!
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  2. #2
    Senior Member topher0805's Avatar
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    You don't have to use limits to find this derivative.

    Recall that:

    \sqrt {x} = x^{\frac {1}{2}}

    So:

    \sqrt {x}+1 = x^{\frac {1}{2}}+1

    So your derivative is just:

    \frac {1}{2\sqrt {x}}
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by sondre View Post
    i'm having a major brain clot,

    the derivative of this function

    f(x)= √x + 1




    using the limit defintion




    thank you so much!
    f(x) = \sqrt{x}+1

    derivative:
    =\lim_{h\to 0} \frac{\sqrt{x+h}+1-\sqrt{x}-1}h

    =\lim_{h\to 0} \frac{\sqrt{x+h}-\sqrt{x}}h * \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}

    =\lim_{h\to 0} \frac{h+x-x}{h(\sqrt{x+h}+\sqrt{x})}

    =\lim_{h\to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}

    =\frac{1}{\lim_{h\to 0} \sqrt{x+h}+\sqrt{x}}

    =\frac{1}{\sqrt{x}+\sqrt{x}}

    =\frac{1}{2\sqrt{x}}
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  4. #4
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    Hello, sondre!

    I assumed that the square root is over the entire x + 1


    Differentate: . f(x)\:= \:\sqrt{x + 1}

    using the limit defintion: . f'(x) \;=\;\lim_{h\to0}\frac{f(x+h) - f(x)}{h}
    There are four steps . . .


    (1) Find f(x+h)\!:\quad f(x+h) \;=\;\sqrt{(x+h)+1}


    (2) Subtract f(x)\!:\quad f(x+h)-f(x) \;=\;\sqrt{x+h+1} - \sqrt{x+1}

    . . Multiply top and bottom by \sqrt{x+h+1} + \sqrt{x+1}

    . . . . \frac{\sqrt{x+h+1} - \sqrt{x+1}}{1}\cdot\frac{\sqrt{x+h+1} + \sqrt{x+1}}{\sqrt{x+h+1} + \sqrt{x+1}}

    . . . . .  =\;\frac{(x+h+1) - (x+1)}{\sqrt{x+h+1} + \sqrt{x+1}} \;=\;\frac{h}{\sqrt{x+h+1}-\sqrt{x+1}}


    (3) Divide by h\!:\quad \frac{1}{h}\cdot\frac{h}{\sqrt{x+h+1} + \sqrt{x+1}} \;=\;\frac{1}{\sqrt{x+h+1} + \sqrt{x+1}}


    (4) Take limit: . \lim_{h\to0}\left[\frac{1}{\sqrt{x +h+1} + \sqrt{x+1}}\right] \;=\;\frac{1}{\sqrt{x+1} + \sqrt{x+1}}


    Therefore: . \boxed{f'(x) \;=\;\frac{1}{2\sqrt{x+1}}}

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  5. #5
    Super Member angel.white's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, sondre!

    I assumed that the square root is over the entire x + 1
    Oh, thats a good call, I hadn't thought of that >.<
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