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Math Help - difference quotient

  1. #1
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    difference quotient

    I am trying to simplify so h does not cause division by zero when h=0. but the algebra is not so obvious to me.

    this is what I try.

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

    also

    \frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h} \;=\;  \frac{1}{h}\cdot \frac{x \sqrt{x+h}-(x+h)\sqrt{x}}{x(x+h)} \;=\; ?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: difference quotient

    Quote Originally Posted by skoker View Post
    I am trying to simplify so h does not cause division by zero when h=0. but the algebra is not so obvious to me.

    this is what I try.

    \frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h} \;=\; \frac{1}{h}\cdot \frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x(x+h)}} \;=\; ?
    \frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h} \;=\; \frac{1}{h}\cdot \frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x(x+h)}}\cdot \frac{\sqrt{x} +\sqrt{x + h}}{\sqrt{x} +\sqrt{x + h}} \;=\; ...
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  3. #3
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    Re: difference quotient

    @Also sprach Zarathustra

    I was suspecting rationalization but did not figure out at what step to use it. thanks.

    so I get this one.

    \frac{\Delta y}{\Delta x} = \frac{-1}{x\sqrt{x+h}+(x+h)\sqrt{x}}

    which gives correct result.
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