Differentiating $\int\cdots \int f(X_1,X_2,\ldots,X_n)\varphi_1(x_1,\theta)\cdots\varphi_n(x_n,\theta)~dx_1\cdots dx_n$ Differentiating:$$\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty f(X_1,X_2,\ldots,X_n)\varphi_1(x_1,\theta)\cdots\varphi_n(x_n,\theta)\,dx_1 \cdots dx_n$$ with respect to $\theta$.
The result is given in one line, (the next one). I do not understand how this is. (Statistics proof)
Anyway the result given being:
$$\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty f(X_1,X_2,\ldots,X_n) \sum_{i=1}^n \left(\frac{\partial}{\partial \theta}\varphi(x_i,\theta)\frac{1}{\varphi(x_i,\theta)}\right) \varphi_1(x_1,\theta)\cdots\varphi_n(x_n,\theta)\,dx_1\cdots dx_n$$

The first question is: Under what circumstances is
$$
\frac \partial {\partial\theta} \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \bullet\bullet\bullet
$$
the same as
$$
\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \frac \partial {\partial\theta} \bullet\bullet\bullet \text{ ?}
$$
The next question is: Why is
$$
\frac \partial {\partial\theta} \varphi_1(x_1,\theta)\cdots\varphi_n(x_n,\theta)
$$
the same thing as
$$
\sum_{i=1}^n \left(\frac{\partial}{\partial \theta}\varphi(x_i,\theta)\frac{1}{\varphi(x_i,\theta)}\right) \varphi_1(x_1,\theta)\cdots\varphi_n(x_n,\theta) \text{ ?}
$$
The answer to that last is the product rule, in this form:
\begin{align}
& \frac d {d\theta} (abcefg) \\[10pt]
= {} & \left( \frac{da}{d\theta}\right) bcefg + a\left( \frac{db}{d\theta}\right) cefg + ab \left( \frac{dc}{d\theta}\right) efg + abc \left( \frac{de}{d\theta}\right) fg + \cdots \\[10pt]
= {} & \left( \frac{da}{d\theta} \cdot \frac 1 a\right) abcefg + \left( \frac{db}{d\theta} \cdot \frac 1 b \right) abcefg + \left( \frac{dc}{d\theta}\cdot \frac 1 c \right) abcefg + \left( \frac{de}{d\theta}\cdot \frac 1 e \right) abcefg + \cdots.
\end{align}
