Answer
The derivative is $\frac{3x\sqrt x+2x}{(2\sqrt x +1)^2}.$
Work Step by Step
Using the quotient rule: $f'(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}
$ $u(x)=x^2; u'(x)=2x$
$v(x)=2\sqrt x +1; v'(x)=\frac{1}{\sqrt x}$
$f'(x)=\dfrac{(2x)(2\sqrt x+1)-(x^2)(\frac{1}{\sqrt x})}{(2\sqrt x+1)^2}$
$=\dfrac{4x\sqrt x+2x-x\sqrt x}{(2\sqrt x +1)^2}=\dfrac{3x\sqrt x+2x}{(2\sqrt x +1)^2}$.
