please differentiate : 3 ( 2t^2 + t ) ^3 +6?

Question

Note : '+t' and '+6' are not parts of the exponents !

Answer 1

9(2t^2 +t)^2 x (4t+1)

Answer 2

if +6 is not a part of the exponent it is a onstant and derivative =0

so you are left with 3 (2t^2+t )^3
derivative of u^3 = 3u^2* du/dt here u = 2t^2+t

so the derivative is 9 (2t^2+t)^2 (4t+1)

Answer 3

9(2t^2+t)^2 (4t+1)

Answer 4

=9(4t+1)(2t^2+1)^2

Answer 5

y=3 (2t^2+t)^3+6
y'=9*(4t+1)( 2t^2+t)^2

i hope that this helps

Answer 6

y = 3(2t² + t)^(3) + 6
let u = 2t² + t
du/dt = 4t + 1
y = 3.(u)³ + 6
dy/du = 9u²
dy/dt = (dy/du).(du/dt)
dy/dt = (9u²).(4t + 1)
dy/dt = 9.(2t² + t)².(4t + 1)
dy/dt = 9.t².(2t + 1)².(4t + 1)