The mgf is given by $\displaystyle M_x(t) = {pe^t \over 1 - (1-p)e^t}$

I have the mean as $\displaystyle M_x'(0) = {1 \over p }\mbox{ and the variance as } M_x''(0) - (M_x'(0))^2 $$\displaystyle \mbox{ and the variance comes to }{1-p \over p^2}$

is this correct??