Can Leibniz integral rule be extended to differentiation under the sigma sign?

Question

To differentiate $\displaystyle M(t)=\sum_i e^{tx_i} P(x_i)$ with respect to $t$, for instance.

Answer 1

Hint: Expand out whatever summation you are working with, i.e. $$\sum\limits_i e^{tx_i}P(x_i)=e^{tx_1}P(x_1)+e^{tx_2}P(x_2)+\cdots$$ Now recall that differentiation is linear and note how the differential operator acts on the rewritten sum: $$\frac{d}{dt}\left(e^{tx_1}P(x_1)+e^{tx_2}P(x_2)+\cdots\right)=\frac{d}{dt}e^{tx_1}P(x_1)+\frac{d}{dt}e^{tx_2}P(x_2)+\cdots$$ What can you now say about the derivative of your orignal sum? $$M'(t)=\frac{d}{dt}\sum_{i}e^{tx_i}P(x_i)=\cdots$$