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Math Help - derivative by first principles

  1. #1
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    derivative by first principles

    Hi all,
    I was wondering how do you find the derivative of the following function by first principles:

    (1 + 2x) ^(1/2)

    When i try to find the derivate by first principles the x cancels out. Well thank you in advance.

    ArTiCk
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  2. #2
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    f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

    f'(x) = \lim_{h \to 0} \frac{\sqrt{1 + 2(x + h)} - \sqrt{1+2x}}{h}

    f'(x) = \lim_{h \to 0} \frac{\sqrt{1 + 2x + 2h} - \sqrt{1+2x}}{h}

    Rationalize the numerator:

    f'(x) = \lim_{h \to 0} \left( \frac{\sqrt{1 + 2x + 2h } - \sqrt{1+2x}}{h} \cdot \frac{\sqrt{1+ 2x + 2h} + \sqrt{1+2x}}{\sqrt{1+ 2x + 2h} + \sqrt{1+2x}}\right)

    f'(x) = \lim_{h \to 0} \frac{1 + 2x + 2h - (1 + 2x)}{h \left(\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x}\right)}

    And simplify. Yes the x's cancel out in the numerator but they're still there in the denominator.
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  3. #3
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    Hello, ArTiCK!

    Find the derivative by first principles: . f(x) \:=\:\sqrt{1 + 2x}

    We want: . f'(x) \;=\;\lim_{h\to0}\frac{f(x+h)-f(x)}{h}


    (1) Find f(x+h)

    . . f(x+h) \:=\:\sqrt{1 + 2(x+h)}


    (2) Subtract f(x)

    . . f(x+h) - f(x) \:=\:\sqrt{1 + 2(x+h)} - \sqrt{1+2x}

    Rationalize: . \frac{\sqrt{1+2(x+h)} - \sqrt{1+2x}}{1}\cdot\frac{\sqrt{1+2(x+h)} + \sqrt{1+2x}}{\sqrt{1+2(x+h)} + \sqrt{1+2x}}

    . . = \;\frac{[1 + 2(x+h)] - [1 + 2x]}{\sqrt{1+2(x+h)} + \sqrt{1 + 2x}} \;= . \frac{1 + 2x + 2h - 1 - 2x}{\sqrt{1+2(x+h)} + \sqrt{1+2x}} \;=\;\frac{2h}{\sqrt{1+2(x+h)} + \sqrt{1+2x}}


    (3) Divide by h.

    . . \frac{f(x+h)-f(x)}{h} \;=\;\frac{2h}{h[\sqrt{1 + 2(h+x)} + \sqrt{1+2x}} \;=\;\frac{2}{\sqrt{1+2(x+h)} + \sqrt{1+2x}}


    (4) Take the limit.

    . . \lim_{h\to0}\frac{f(x+h)-f(x)}{h} \;=\;\lim_{x\to0}\frac{2}{\sqrt{1+2(x+h)} + \sqrt{1+2x}} \;=\;\frac{2}{\sqrt{1+2x}\cdot\sqrt{1+2x}}

    . . . . . = \;\frac{2}{2\sqrt{1+2x}} \;=\;\frac{1}{\sqrt{1+2x}}


    Therefore: . \frac{d}{dx}\left(\sqrt{1+2x}\right) \;=\;\frac{1}{\sqrt{1+2x}}

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