This is an exercise from Khan Academy, using the quotient rule. What are the steps for finding the derivative of this function: $$f(x)=\frac{\sin(x)-\sqrt{x}\cos(x) }{2\sqrt{x}\sin^2(x)}$$
To this: In the next part a "$2x$" appears replacing "$\sqrt{x}$" in the numerator. What is the step there? $$\frac{\sin(x)-2x\cos(x)}{2\sqrt{x}\sin^2(x)}$$
Ok I think I see where you are talking about. He has $\frac{\frac{1}{2\sqrt{x}}\cdot \sin(x)-\sqrt{x} \cdot \cos(x)}{\sin^2(x)}$
To clean this up. He multiplies top and bottom by $2 \sqrt{x}$ giving:
$\frac{\color{red}{2 \sqrt{x}}\frac{1}{2\sqrt{x}}\cdot \sin(x)-\color{red}{2 \sqrt{x}}\sqrt{x} \cdot \cos(x)}{\color{red}{2 \sqrt{x}}\sin^2(x)}$
Doing a little simplifying gives:
$\frac{ \sin(x)-2 x \cos(x)}{ 2 \sqrt{x} \sin^2(x)}$
The following is a print screen of khan's site: