# difference quotient

• Jul 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

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

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

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