Question

If $u=\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|$, then $\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}$ is equal to

• A. $0$
• B. $1$
• C. $2u$
• D. $3u$

Answer 1

The correct option is A $0$

The given information is: $u=\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|$

Now we calculate $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$ and $\frac{\partial u}{\partial z}$

$\Rightarrow \frac{\partial u}{\partial x}=\left|\begin{array}{ccc}0& 1& 2x\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|$

$\Rightarrow \frac{\partial u}{\partial y}=\left|\begin{array}{ccc}1& x& x^2\\ 0& 1& 2y\\ 1& z& z^2\end{array}\right|$

$\Rightarrow \frac{\partial u}{\partial z}=\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 0& 1& 2z\end{array}\right|$

Adding all the values we get,

$\Rightarrow \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}=\left|\begin{array}{ccc}0& 1& 2x\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|+\left|\begin{array}{ccc}1& x& x^2\\ 0& 1& 2y\\ 1& z& z^2\end{array}\right|+\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 0& 1& 2z\end{array}\right|$

Expanding the determinants along the 1sr row we get,

$\Rightarrow \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}=(y^2-z^2)+(2xz-2yx)+(z^2-2yz)+2yx-x^2+(2yz-y^2)-2xz+x^2$

$\Rightarrow \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}=0$