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Math Help - Derivative (1st Principles)

  1. #1
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    Derivative (1st Principles)

    Find the derivative of the function using the definition of derivative.

    1) x + sqrt(x)

    2) 1 / sqrt(x)

    ---------------
    1) ((x+h) + sqrt(x+h) - [x + sqrt(x)])/h
    I'm stuck here, not sure what to do?

    2) [(1/sqrt(x+h)) - (1/sqrt(x))]/h
    Not sure what to do here either?
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  2. #2
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    Hello, millerst!

    This is a test of your algebraic ability and stamina.
    I'll do the first one (it's harder).


    Find the derivative of the function using the definition of derivative.

    . . 1)\;\;f(x) \:=\: x + \sqrt{x}
    Definition of derivative: . f\:\!'(x) \;=\;\lim_{h\to0} \frac{f(x+h) - f(x)}{h}


    My recommendation: divide the procedure into four steps:

    . . [1] Find f(x+h) . . . replace x with x+h ... and simplify.

    . . [2] Subtract f(x) . . . subtract the original function ... and simplify.

    . . [3] Divide by h . . .factor and simplify.

    . . [4] Take the limit as h\to0.


    Here we go . . .


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


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

    . . . \text{Multiply by: }\,\frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h}+\sqrt{x}}:\quad \frac{h+\sqrt{x+h}-\sqrt{x}}{1}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{  x+h}+\sqrt{x}}

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

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

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


    [3] Divide by h\!:\quad \frac{1}{{\color{red}\rlap{/}}h}\cdot\frac{{\color{red}\rlap{/}}h\left(\sqrt{x+h} + \sqrt{x} + 1\right)}{\sqrt{x+h} + \sqrt{x}} \:=\:\frac{\sqrt{x+h}+\sqrt{x}+1}{\sqrt{x+h}+\sqrt  {x}}


    [4] Take the limit:

    . . \lim_{h\to0}\frac{\sqrt{x+h} + \sqrt{x} + 1}{\sqrt{x+h} + \sqrt{x}} \;=\;\frac{\sqrt{x} + \sqrt{x} + 1}{\sqrt{x} + \sqrt{x}} \;= \;\frac{2\sqrt{x} + 1}{2\sqrt{x}} \;=\;\frac{2\sqrt{x}}{2\sqrt{x}} + \frac{1}{2\sqrt{x}}


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



    I need a nap . . .
    .
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  3. #3
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    Hi, thanks for the reply. I'm confused at this step:

    <br /> <br />
= \;\frac{h(\sqrt{x+h} + \sqrt{x}) + h}{\sqrt{x+h} + \sqrt{x}} \;=\; \frac{h\,(\sqrt{x+h}+\sqrt{x} + 1)}{\sqrt{x+h}+\sqrt{x}}<br />


    How did the + h, go into the bracket and become + 1?
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  4. #4
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    Hello, millerst

    I'm confused at this step:

    \frac{h(\sqrt{x+h} + \sqrt{x}) + h}{\sqrt{x+h} + \sqrt{x}} \;=\; \frac{h\,(\sqrt{x+h}+\sqrt{x} + 1)}{\sqrt{x+h}+\sqrt{x}}<br />

    How did the + h, go into the bracket and become + 1?

    We factor out the h.


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  5. #5
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    Wow, I can't believe I didn't see that. Thanks so much.
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