Let X~Exponential(1).
I know that the moment-generating function of an exponential distribution is defined as $(1-t\lambda)^{-1}$. And hence $E[e^{tx}]=(1-t)^{-1}$.
But what is $E[Xe^{tx}]$? Would that be the first moment of the moment-generating function?
Assuming your parameter $\lambda$ is taken to be the expected value of $X$, $X$ has probability density $$f_{X} (x) = \frac{1}{\lambda} \exp\left(-\frac{x}{\lambda}\right).$$ So the expected value $E\left[Xe^{tX}\right]$ is just the integral $$\frac{1}{\lambda} \int_{0}^{\infty} x e^{tx} e^{-x/\lambda} \ dx$$ which reduces to some integration by parts. Assuming that $t < 1/\lambda$, $$E\left[Xe^{tX}\right] = \lim_{y \to \infty} - \frac{1}{1 - \lambda t} ye^{-( 1/\lambda - t)y} + \frac{1}{1 - \lambda t}\int_{0}^{\infty} e^{(t - 1/\lambda)x} \ dx$$ $$= \frac{\lambda}{(1 - \lambda t)^2}.$$
Alternatively, $$E\left[X e^{tX} \right] = E[X] + t E[X^2] + \frac{1}{2!} t^2 E[X^3] + \dots$$ $$= \frac{d}{dt} \left(\sum_{n=0}^{\infty} \frac{1}{n!} t^{n} E[X^n]\right)$$ $$= M_{X}'(t),$$ where $M_{X} (t)$ is your MGF.