Answer
For every $p\in\{1,2,\ldots,n\}$ the derivative is
$$\frac{\partial}{\partial x_p}u=p\cos(x_1+\ldots+(p-1)x_{p-1}+px_p+(p+1)x_{p+1}+\ldots+nx_n).$$
Work Step by Step
We will choose some $p\in\{1,2,\ldots,n\}$ and differentiate with respect to $x_p$ while holding all other $x_i$-s where $i\neq p$ constant:
$$\frac{\partial}{\partial x_p}u=\\\frac{\partial}{\partial x_p}\sin(x_1+\ldots+(p-1)x_{p-1}+px_p+(p+1)x_{p+1}+\ldots+nx_n)=\\
\cos(x_1+\ldots+(p-1)x_{p-1}+px_p+(p+1)x_{p+1}+\ldots+nx_n)\times\\
\times\frac{\partial}{\partial x_p}(x_1+\ldots+(p-1)x_{p-1}+px_p+(p+1)x_{p+1}+\ldots+nx_n)=\\
\cos(x_1+\ldots+(p-1)x_{p-1}+px_p+(p+1)x_{p+1}+\ldots+nx_n)\times\\
\times\left(\underbrace{0+\ldots 0+\frac{\partial}{\partial x_p}(px_p)}+0+\ldots0\right)=\\
p\cos(x_1+\ldots+(p-1)x_{p-1}+px_p+(p+1)x_{p+1}+\ldots+nx_n).$$
This formula works for every $p\in\{1,2,\ldots,n\}$ so if we want to get a particular first partial derivative we just put the corresponding value for $p$ (e.g. if we want $\partial u/\partial x_3$ we put $p=3$).