Question: Differentiate:
$$\frac{2x-1}{(x^2+5)^{0.5}}$$

With the quotient rule 
$\dfrac{v\cdot \frac{du}{dx}-u\cdot\frac{dv}{dx}}{v^2}$

I keep coming to
$\frac{[2(x^2+5)^{0.5}-0.5(2x-1)(x^2+5)^{-0.5}]}{ (x^2+5)}$

With the product rule
$u\cdot\frac{dv}{dx}+v\frac{du}{dx}$
 I made it $(2x-1)(x^2+5)^{-0.5}$ and then

$(2x-1)(-0.5(x^2+5)^{-1.5})+2(x^2+5)^{-0.5}$
I don't know which step I am missing here. 
All help appreciated.

Answer: 
Let $$f(x) = \frac{2x-1}{\sqrt{x^2+5}}$$

* By product rule: $$f’(x) = (2x-1) [\frac{-x}{(x^2+5)^{\frac32}}] + \frac{2}{\sqrt{x^2+5}}$$ $$= \frac{x + 10}{(x^2+5)^{\frac32}}$$
* By quotient rule: $$f’(x) = \frac{2\sqrt{x^2+5}-(2x-1)\frac{x}{\sqrt{x^2+5}}}{x^2+5}$$ $$=\frac{2(x^2+5)-2x^2+x}{(x^2+5)^{\frac32}}$$ $$=\frac{x+10}{(x^2+5)^{\frac32}}$$
$$\left(\frac{2x-1}{\sqrt{x^2+5}}\right)'=\frac{2\sqrt{x^2+5}-\frac{2x(2x-1)}{2\sqrt{x^2+5}}}{x^2+5}=\frac{2(x^2+5)-2x^2+x}{\sqrt{(x^2+5)^3}}=\frac{x+10}{\sqrt{(x^2+5)^3}}$$