I can't understand the algebraic simplification below

Question

I can't understand the last step of simplification in the following algebraic expression.

I am aware that $\sqrt{3}=3^{\frac{1}{2}}$. Can't see how they get the $6$ in the denominator, and how $\sqrt{3}$ gets to the numerator. Any help will be appreciated.

p.s. this is from a mathematics text book. last step of a derivative calculation.

$=\frac{1}{2\sqrt{3}} =\frac{\sqrt{3}}{6}$

Answer 1

Consider $$ \frac{1}{2 \sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} \frac{1}{2 \sqrt{3}} = \frac{\sqrt{3}}{2 \sqrt{3} \sqrt{3}} = \frac{\sqrt{3}}{6}$$ As can be seen I have multiplied the fraction by $1 = \sqrt{3} \ / \sqrt{3}$.

Answer 2

Notice, we have $$\frac{1}{2\sqrt{3}}$$ $$=\frac{1}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$$ $$ =\frac{\sqrt 3}{2\sqrt{3}\sqrt 3}$$ $$=\frac{\sqrt 3}{2\times 3}$$ $$=\frac{\sqrt 3}{6}$$

Answer 3

$$\frac{1}{2\sqrt3}=\frac{1}{2\sqrt3}\times1=\frac{1}{2\sqrt3}\times\frac{\sqrt3}{\sqrt3}=\frac{1\times\sqrt3}{2\sqrt3\times\sqrt3}=\frac{\sqrt3}{2\times3}=\frac{\sqrt3}{6}$$

Answer 4

The two tricks to get square roots out of fraction denominators are:

1. If you have a fraction which looks like: $$A=\frac{x}{a+\sqrt{b}}$$ then $$A=1\cdot A = \frac{a-\sqrt b}{a-\sqrt b}\cdot \frac{x}{a+\sqrt b} = \frac{x(a-\sqrt b)}{(a-\sqrt b)(a+\sqrt b)} - \frac{x(a-\sqrt b)}{a^2-b}$$

2. If you have a fraction which looks like $$B=\frac{x}{\sqrt c}$$ then $$B=1\cdot b = \frac{\sqrt c}{\sqrt c}\cdot \frac{x}{\sqrt c} = \frac{x\sqrt c}{\sqrt c \sqrt c} = \frac{x\sqrt c}{c}$$