Question

Derivada de
$y= \sqrt{sen6x}$

Answer 1

$y=\sqrt{\sin(6x)}$
$u=6x$
$y=\sqrt{\sin(u)}$
$v=\sin(u)$
$y=\sqrt{v}$

$y'=\frac{1}{2\sqrt{v}}\cdot u'\cdot v'$

$u'=6$

$v'=\cos(u)$

$y'=\frac{1}{2\sqrt{v}}\cdot 6\cdot \cos(u)$

$y'=\frac{1}{\sqrt{\sin(6x)}}\cdot 3\cdot \cos(6x)$

Answer 2

$\sf \displaystyle y=\sqrt{sen\:6\:x}\\\\\\\frac{d}{dx}\left(\sqrt{\sin \left(6x\right)}\right)\\\\\\=\frac{1}{2\sqrt{\sin \left(6x\right)}}\frac{d}{dx}\left(\sin \left(6x\right)\right)\\\\\\\to \boxed{\sf =\frac{1}{2\sqrt{\sin \left(6x\right)}}\cos \left(6x\right)\cdot \:6}$