The problem: Let $ X_1,X_2,\cdots ,X_n $ be a random sample from the uniform distribution with pdf $f(x;\theta _1,\theta _2)=\frac{1}{2\theta _2} , \theta _1-\theta _2<x<\theta _1+\theta , $ where$ -\infty <\theta _1<\infty ,0<\theta _2<\infty $ and the pdf is equal to zero elsewhere.
Show that $Y_1=min(X_i) $ and $Y_n=max(X_i)$, the joint sufficient statistic for $\theta _1 $ and $\theta _2$ are complete.
My question: After some calculations, I have found that showing the completeness is equivalent to showing the statement below.
$\int_{\theta _1-\theta _2}^{\theta _1+\theta _2} \int_{\theta _1-\theta_2}^{y_n} u(y_1,y_n)(y_n-y_1)^{n-2}/(2\theta _2)^{n}dy_1dy_n=0 $ for all $-\infty <\theta _1<\infty ,0<\theta _2<\infty$ $\Rightarrow u(y_1,y_n)=0$
I looked at the solution and it was written like this: Multiply by $(2\theta _n)^{n}$ and differentiate first with respect to $\theta_1$ and then $\theta_2$. This finally yieds $u(\theta _1-\theta _2,\theta _1+\theta _2)=0 $ , for all $(\theta _1,\theta _2)$, which implies that $u(y_1,y_n)=0$.
I'm stuck in differentiating the double integrals. I have seen differentiaing the single integral using the fundamental theorem of calculus, but in this case , I don't know how to differentiate.