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Thread: 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:

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

    So:

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

    So your derivative is just:

    $\displaystyle \frac {1}{2\sqrt {x}}$
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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!
    $\displaystyle f(x) = \sqrt{x}+1$

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

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

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

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

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

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

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

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


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

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


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


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

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

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

    . . . . . $\displaystyle =\;\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 $\displaystyle 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: . $\displaystyle \lim_{h\to0}\left[\frac{1}{\sqrt{x +h+1} + \sqrt{x+1}}\right] \;=\;\frac{1}{\sqrt{x+1} + \sqrt{x+1}}$


    Therefore: .$\displaystyle \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 $\displaystyle x + 1$
    Oh, thats a good call, I hadn't thought of that >.<
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