Question

Помогите найти производную функции

Answer 1

( 8xtg3x-(4x²-2)*3\cos²3x ) \ tg²3x =  

( 8xsin3x\cos3x - (12x²-6)\cox²3x) \ tg²3x =

(4xsin6x-12x²+6)  \  sin²3x

Answer 2

$\frac{4x^2-2}{tg3x} = \frac{8xtg3x-\frac{12x^2-6 }{cos^23x} }{tg^23x} \\ \\ (tg3x)'=3(\frac{sin3x}{cos3x})'=3* \frac{cos^23x+sin^23x}{cos^23x} = \frac{3}{cos^23x}$