Question

2 задание,производную найти​

Answer 1

y=sin(x)+2x³-4ln(x)

y'= cos(x)+6x²- (4/x)

Answer 2

$y=sinx+2x^3-4\, lnx\\\\y'=cosx+2\cdot 3x^2-4\cdot \frac{1}{x}=cosx+6x^2-\frac{4}{x}$