Question

Найти производную функции у=(cos2x)^sin3x

Answer 1

$\y=(cos2x)^{sin3x}\ y=(e^{ln cos 2x})^{sin 3x}\ y=e^{sin 3xln cos 2x}\ y'=e^{sin 3xln cos 2x}cdot3cos 3xln cos 2x+sin3xcdotfrac{1}{cos 2x}cdot(-2sin2x)\ y'=(cos2x)^{sin3x}3cos3xlncos2x-2sin3xtan2x\ y'=cos^{sin 3x}2x3cos3xlncos2x-2sin3xtan2x$