How does the difference quotient with a square root in the numerator end up with square roots in the denominator?

Question

I don't understand when

I apply the difference quotient to: $f(x) = \sqrt{x} $ , to get:

$$\frac{\sqrt{x+h} - \sqrt{x}}{h}$$

To simplify it.. How does it end up like this?:

$$\frac{x + h - x}{h \sqrt{x+h} + \sqrt{x}}= \frac{1}{\ 2\sqrt{x}}$$

How do the sqrt's work when moving them from the numerator to the denominator?

Thanks.

Answer 1

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

So by letting h go to 0 we get

$\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}=\frac{1}{2\sqrt{x}}$