Let $x$ $=(x_1, x_2, x_3)'$, and $u(x)=3x_1^3 -2x_1^2x_2+4x_3$. Find$\frac{\partial u}{\partial x}$.
So far I have done a simple differentiation and got
$$ \frac{\partial u}{\partial x} = \begin{pmatrix} 9x_1^2-4x_1x_2+4x_3 \\ 3x_1^3-2x_1^2+4x_3 \\ 3x_1^3-2x_1^2x_2+4 \end{pmatrix}$$
I have doubts and I think I am under-thinking this. Does anyone have any advice if I am missing anything?
Thank you very much in advance!
We we compute $\frac{\partial u}{\partial x_1}$, we view the other two variables $x_2, x_3$ as constants. So $$\frac{\partial u}{\partial x_1}=\frac{\partial}{\partial x_1}(3x_1^3)-\frac{\partial}{\partial x_1}(2x_1^2x_2)+\frac{\partial}{\partial x_1}(4x_3)=9x_1^2-4x_1x_2+\color{red}{0}.$$ For the same reason, we have $$\frac{\partial u}{\partial x_2}=\frac{\partial}{\partial x_2}(3x_1^3)-\frac{\partial}{\partial x_2}(2x_1^2x_2)+\frac{\partial}{\partial x_2}(4x_3)=\color{red}{0}-2x_1^2+\color{red}{0},$$ $$\frac{\partial u}{\partial x_3}=\frac{\partial}{\partial x_3}(3x_1^3)-\frac{\partial}{\partial x_3}(2x_1^2x_2)+\frac{\partial}{\partial x_3}(4x_3)=\color{red}{0}+\color{red}{0}+4.$$
Therefore, $$ \frac{\partial u}{\partial x}\text{(It is more common to denote it by $\nabla u$)} = \begin{pmatrix} 9x_1^2-4x_1x_2 \\ -2x_1^2 \\ 4 \end{pmatrix}.$$