I don't know how to get the logarithm of this
I have $S(t)=\log(M_{X}(t))$.
Show that $$ \tfrac{d~}{dt}S(t)|_{t=0}=E(X) $$ and $$ \tfrac{d^{2}~}{dt~^{2}}S(t)|_{t=0}=Var(X) $$ with $m_{X}(t)=E(\mathrm e^{tx})$
My question is how I can calculate the derivative of this expression. $$\tfrac{d~}{dt}S(t)|_{t=0}= \tfrac{d~}{dt}\log(M_{X}(t))|_{t=0}=\tfrac{d~}{dt}\log(E(\mathrm e^{tX}))|_{t=0}$$
Use these: $$\begin{align}\dfrac{\mathrm d \log( g(t))}{\mathrm d t}&=\dfrac{1}{g(t)}\cdot\dfrac{\mathrm d g(t)}{\mathrm d t}&&\text{via chain rule (presuming }\log\text{ is base e)}\\[2ex]\dfrac{\mathrm d(1/g(t))}{\mathrm d t}&=\dfrac{-1}{g(t)^2}\cdot\dfrac{\mathrm d g(t)}{\mathrm d t}&&\text{via chain rule}\\[2ex]\dfrac{\mathrm d~}{\mathrm d t}\mathsf E(h(t,X))&=\mathsf E\left(\dfrac{\partial~}{\partial t}h(t,X)\right)&&\text{via definition of expectation}\end{align}$$