1. ## simplified derivative?

$

f(x) = 2x^2sin3x

$

derivative

$

f '(x) = 4xsin3x + 2x^2(3cos3x)

$

but simplified its

$

f '(x) = 2x(2sin3x + 3xcos3x)

$

how did they manage to simplify it like that? thank u

2. $f '(x) = 4xsin3x + 2x^2(3cos3x) = 4xsin3x + 6x^2cos3x$

Now factor out 2x