yes

( f( x )*g( x ) ) ' = f ' ( x )*g( x ) + g ' ( x )*f( x )

you can put:

if f( x ) = ( x + 20 )^4 => f ' ( x ) = 4*( x + 20 )^3

g( x ) = ( x - 30 )^3 => g ' ( x ) = 3*( x - 30 )^2

and if I use that in the formula above i will get the same thing as you