summation - Series with Binomial Coefficients

I need to get a closed form for this series $$\sum_{x=0}^{\infty} x {z \choose x} \lambda ^ x \mu^{z-x}$$

I know that that $\sum_{x=0}^{\infty} {z \choose x} \lambda ^ x \mu^{z-x} = (\lambda + \mu)^z$ (formally) and I feel that I am supposed to proceed from here by differentiation, but I do not know how.

Answer

Let

$$\sum_{x=0}^{\infty} {z \choose x} \lambda ^ x \mu^{z-x}=(\lambda + \mu)^z$$
by differentiation about $\lambda$:
$$\sum_{x=0}^{\infty} x {z \choose x} \lambda ^ {x-1} \mu^{z-x}=z(\lambda + \mu)^{z-1}$$
multiple two sides with $\lambda$
$$\sum_{x=0}^{\infty} x {z \choose x} \lambda ^ x \mu^{z-x}=z\lambda(\lambda + \mu)^{z-1}$$