difference quotient

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July 6th 2011, 02:42 PM
skoker

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

$\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h} \;=\; \frac{1}{h}\cdot \frac{x \sqrt{x+h}-(x+h)\sqrt{x}}{x(x+h)} \;=\; ?$
July 6th 2011, 03:21 PM
Also sprach Zarathustra

Re: difference quotient

Quote:

Originally Posted by skoker

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.

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

$\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h} \;=\; \frac{1}{h}\cdot \frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x(x+h)}}\cdot \frac{\sqrt{x} +\sqrt{x + h}}{\sqrt{x} +\sqrt{x + h}} \;=\; ...$
July 6th 2011, 05:24 PM
skoker

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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