Most of our practice problems ask to find MGFs but never really to use them. Thus I 'm confused by why t was set to equal 0 in this question:
There are variables $Y_1 = Z$ and $Y_2 = Z^2$ where $Z$ ~$N(0,1)$.
The mgf of $ N(0,1)$ is $m(t)=e^{t^2 /2}$. Find $E(Y_1 Y_2)$. The solutions involve taking the third derivative of the mgf and setting t =0. Since $m^3(t)=t^3 e^{t^2 /2}+2te^{t^2 /2}+te^{t^2 /2}$ the answer was zero.
Why was t set to zero?
Here $Y_1Y_2=Z^3$ and the moment generating function of $Z$ is $M_Z(t) = e^{t^2/2}$. Since $M_z$ is infinitely differentiable and $Z$ is symmetric about zero (i.e. $f_Z(z) = f_Z(-z)$ for all $z\in\mathbb R$), it follows that all odd moments (in particular $\mathbb E[Z^3]$) are zero. No need to actually differentiate here.