Answer
$u_{x_1}=\cos{(x_1+2x_2+...+nx_n)}$, $u_{x_2}=2\cos{(x_1+2x_2+...+nx_n)}$,,,$u_{x_n}=n\cos{(x_1+2x_2+...+nx_n)}$.
Work Step by Step
$u=\sin{(x_1+2x_2+...+nx_n)}$
In order to find $u_{x_1}$ we treat all other variables as constants and differentiate with respect to $x_1$.
$u_{x_1}=\cos{(x_1+2x_2+...+nx_n)}$
Analogously:
$u_{x_2}=2\cos{(x_1+2x_2+...+nx_n)}$
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$u_{x_n}=n\cos{(x_1+2x_2+...+nx_n)}$
