I have a function which is defined as \begin{equation} P(x_i,y_i)=\iint_{(x-x_i)^2+(y-y_i)^2\leq\delta}f(x,y,x_i,y_i) \, dx \, dy. \end{equation} How to calculate the derivative with respective $x_i$? Is it enough to write \begin{equation} \frac{\partial P(x_i,y_i)}{\partial x_i}=\iint_{(x-x_i)^2+(y-y_i)^2 \leq \delta} \frac{\partial f(x,y,x_i,y_i)}{\partial x_i} \, dx\,dy. \end{equation}
Since the boundary is also a function of $x_{i}$, do we have to add another term like the Leibniz rule for differentiating an integral?
Enforcing the substitution $\xi=x-x_i$ and $\eta =y-y_i$, we have
$$P(x_i,y_i)=\underbrace{\iint_{\xi^2+\eta^2\le \delta^2}}_{\text{Independent of}\,x_i,y_i}f(\underbrace{\xi+x_i,\eta+y_i}_{\text{Dependent on}\,x_i,y_i},x_i,y_i)\,d\xi\,d\eta$$
Differentiating with respect to $x_i$ yields
$$\begin{align} \frac{\partial P(x_i,y_i)}{\partial x_i}&=\iint_{\xi^2+\eta^2\le \delta^2}\left(f_1(\xi+x_i,\eta+y_i,x_i,y_i)+f_3(\xi+x_i,\eta+y_i,x_i,y_i)\right)\,d\xi\,d\eta\\\\ &=\iint_{\xi^2+\eta^2\le \delta^2} \frac{\partial f(\xi+x_i,\eta+y_i,x_i,y_i)}{\partial \xi}\,d\xi\,d\eta\\\\ &+\iint_{\xi^2+\eta^2\le \delta^2}f_3(\xi+x_i,\eta+y_i,x_i,y_i)\,d\xi\,d\eta\\\\ &=\underbrace{\int_{-\delta}^{\delta} \left(f\left(x_i+\sqrt{\delta^2-\eta^2},\eta+y_i,x_i,y_i\right)-f\left(x_i-\sqrt{\delta^2-\eta^2},\eta+y_i,x_i,y_i\right)\right)\,d\eta}_{\text{"Boundary Term"}}\\\\ &+\iint_{\xi^2+\eta^2\le \delta^2}f_3(\xi+x_i,\eta+y_i,x_i,y_i)\,d\xi\,d\eta \end{align}$$
Now, finish by carrying out the inner integral of the first integral on the right-hand side of $(1)$.
Alternatively, we can write the original integral as
$$\begin{align} \frac{\partial P(x_i,y_i)}{\partial x_i}&=\frac{\partial }{\partial x_i}\int_{y_i-\delta}^{y_i+\delta}\int_{x_i-\sqrt{\delta^2-(y-y_i)^2}}^{x_i+\sqrt{\delta^2-(y-y_i)^2}}f(x,y,x_i,y_i)\,dx\,dy\\\\ &=\int_{y_i-\delta}^{y_i+\delta} \left( f\left(x_i+\sqrt{\delta^2-(y-y_i)^2},y,x_i,y_i\right)-f\left(x_i-\sqrt{\delta^2-(y-y_i)^2},y,x_i,y_i\right) \right)\,dy\\\\ &+\iint_{(x-x_i)^2+(y-y_i)^2\le \delta^2}\frac{\partial f(x,y,x_i,y_i)}{\partial x_i}\,dx\,dy \end{align}$$