Question

VT=2t^2+3t+1 quando t0=5
derivada

Answer 1

$V(t) = 2t^2 +3t+1$

a derivada
$V'(t) =2*2t^{2-1}+3*t^{1-1} + 0\\\\ V'(t) =4t+3$

a derivada quando t=5
$V'(5) = 4*5 + 3\\\\V'(5) = 20+3 = 23$

Answer 2

VT = 2t²+3t+1
derivda :
dVT = 2t²+3t+1 dt
       = 2.2t + 3
       = 4t + 3,

t0 = 5
   dVT = 4.5 + 3
          = 20 + 3
          = 23