Question

Determine the second derivative evaluated at t=0

Answer 1

If

$M(t) = e^{3(e^t-1)}$

then using the chain rule, we have

$M'(t) = e^{3(e^t-1)} \times 3e^t = 3e^{3(e^t-1)+t}$

$M''(t) = 3e^{3(e^t-1)+t} \times (3e^t+1)$

Then

$M''(0) = 3e^{3(e^0-1)+0} \times (3e^0+1) = 3e^0 \times (3+1) = \boxed{12}$