Question

найти производную функции:
y=4√sin2x​

Answer 1

$y=4\sqrt{sin2x}\\\\\sqrt{u}=\frac{1}{2\sqrt{u}}\cdot u'\\\\y'=4\cdot \frac{1}{2\sqrt{sin2x}}\cdot (sin2x)'=\; \; [\; (sinu)'=cosu\cdot u'\; ]=\\\\=\frac{2}{\sqrt{sin2x}}\cdot cos2x\cdot (2x)'=\frac{2}{\sqrt{sin2x}}\cdot cos2x\cdot 2=\frac{4\, cos2x}{\sqrt{sin2x}}$