1. ## Derivative...

Original: If f(x)=2 (Root(x))ln(x), find f'(x).

u' = 1/2*2 (x^(1/2-1) = x^-1/2= 1/x^1/2
v' ln(x) = 1/x

Did the product rule.... and got

My answer was 1/x^1/2*(ln(x)) + 1/x*2(x^1/2)

I typically get these types of problems right but I just cant figure out what I did wrong... If anyone can spot my error help would be greatly appreciated!!

2. Originally Posted by sitdownson
Original: If f(x)=2 (Root(x))ln(x), find f'(x).

u' = 1/2*2 (x^(1/2-1) = x^-1/2= 1/x^1/2
v' ln(x) = 1/x

Did the product rule.... and got

My answer was 1/x^1/2*(ln(x)) + 1/x*2(x^1/2)

I typically get these types of problems right but I just cant figure out what I did wrong... If anyone can spot my error help would be greatly appreciated!!
$f(x)=2\sqrt{x}\ lnx=2uv$

$\displaystyle\ u=\sqrt{x}\Rightarrow\ u'=\frac{1}{2\sqrt{x}}$

$\displaystyle\ v=lnx\Rightarrow\ v'=\frac{1}{x}$

$\displaystyle\ 2\frac{d}{dx}(uv)=2\left(vu'+uv'\right)=\frac{lnx} {\sqrt{x}}+\frac{2\sqrt{x}}{x}$

which is what you got.
It can be be simplified a little to

$\displaystyle\ \frac{lnx}{\sqrt{x}}+\frac{2\sqrt{x}}{\sqrt{x}\sqr t{x}}=\frac{lnx+2}{\sqrt{x}}$