A particle's position with respect to time as it moves along a coordinate axis is given by a cubic function, p(t) = t + 5t + 4t + 3

The position = p(t) = t³+ 5t²+ 4t + 3

To find the velocity, get the derivative of the position which is

velocity: p'(t) = 3t²+ 10t + 4

To find the acceleration, find the derivative of the velocity, which is

Acceleration: p"(t) = 6t + 10

If the particle's acceleration at time t = 3 then we plug in -3 into what we have for the acceleration to get the following:

p(t) = 6t+10 = p(-3)= 6(-3)+10 = -8

HOWEVER!! ..Here is my question: If our equation would have been 3(t^3 + 16t^2 + 4t - 26) instead, would this significantly affect the derivative? What rule do you have to follow now to take the derivative?

How would I solve this?