Silly algebraic or arithmetic issue in solving derivatives using the quotient rule

[Jimmykerrpaisley]

I am doing Khan Academy's calculus course. I am on derivatives and the quotient rule d/dx(f/g)
=(f'(x)*g(x)-f(x)*g'(x))÷g^2 ................................edited
I thought I was doing okay. The exercises seemed pretty simple, similar to the product rule except you subtract instead of All good..... so far. Now I get this.....
Ln(x)/sq.root(x) sorry can't find radical on keyboard
I simplified it down to this 1/sq.root (x) - ln(x)/2x(sq.root(x)
Khan Ac says that it can be simplified further into
2-ln(x)/2xsq.root(x). Eh???? Where did that top 2 come from.
I know I'm being thick. I'm doing something daft and either my algebra or arithmetic is failing. I'm really worried because I have already spent the past year going through Algebra 1 and 2. It was quite hard, especially logs and quotient stuff, although Trig just blew my mind, but never mind that. I am not mathematical naturally, please help this old man try and learn. Thank you so much.

 

[Harry_the_cat]

\(\displaystyle f(x) =ln (x) => f '(x) =\frac{1}{x}\)

\(\displaystyle g(x) = \sqrt{x} => g '(x) =\frac{1}{2\sqrt{x}}\)

Using quotient rule:

\(\displaystyle y' = \frac{\frac{1}{\sqrt{x}} - \frac{ln(x)}{2\sqrt{x}}}{x}\) as you had

\(\displaystyle = \frac{\frac{2}{2\sqrt{x}} - \frac{ln(x)}{2\sqrt{x}}}{x}\) to get common denominator on top

\(\displaystyle = \frac{2-ln(x)}{2x\sqrt{x}}\)