Is this derivative correct?

Question

I'm newbie at Calculus, so I'm doing some exercises of derivates, I know by the formula:

$f(x) = \sqrt u$

$\frac {df(x)}{dx} = \frac{u'}{2 \sqrt u}$

that the derivate of the next function is:

$f(x) = \sqrt{4x^3-2x^2 +4}$

$f'(x) = \frac {12x^2-4x}{2 \sqrt{4x^3-2x^2+4}}$

but if I do WolframAlpha derivates it, it returns:

$\frac{x(3x-1)}{\sqrt{x^3 - \frac{x^2}{2}+1}}$

I know to get this result I have to divide both numerator and denominator for $4$ and then factorize $x$ in the numerator, what I can't understand is where is the $2$ that is in my derivate on the denominator.

Thank you in advance :)

Answer 1

\begin{align} \frac {12x^2-4x}{2\sqrt{4x^3-2x^2+4}}&= \frac{6x^2-2x}{\sqrt{4x^3-2x^2+4}}\\ &= \frac{6x^2-2x}{2\sqrt{x^3-\frac{x^2}{2}+1}}\\ &= \frac{3x^2-x}{\sqrt{x^3-\frac{x^2}{2}+1}}\\ &= \frac{x(3x-1)}{\sqrt{x^3 - \frac{x^2}{2}+1}} \end{align}

Answer 2

You can factor a 2 out of the numerator of the derivative, and this will cancel with the 2 in the denominator.

Answer 3

In fact, they are equal.

$f'(x) = \frac {12x^2-4x}{2 \sqrt{4x^3-2x^2+4}}$

both divide 4, we can get $\frac{x(3x-1)}{\sqrt{x^3 - \frac{x^2}{2}+1}}$.