Question

яка похыдна выд y=sqrt{x}(tg6x)

Answer 1

$y'=(sqrt{x}cdot tg6x)'=(sqrt{x})'cdot tg6x + sqrt{x}cdot (tg6x)'=$

 

$frac{1}{2sqrt{x}}cdot tg6x + sqrt{x}cdot frac{1}{cos^2x}cdot(6x)'=frac{tg6x }{2sqrt{x}} + frac{6sqrt{x}}{cos^2x}$