I came across this question on Khan Academy:
let $f(x, y, z) = \sqrt{xyz}$, find $\frac{\partial f}{\partial x}\bigg\rvert_{(-1, -1, 4)}$
My first intuition was to use the power rule:
$$ \frac{\partial f}{\partial x}=\sqrt{yz}\cdot\frac{\partial}{\partial x}\big(\sqrt{x}\big)=\frac{\sqrt{yz}}{2\sqrt{x}} $$
$$ \frac{\partial f}{\partial x}\bigg\rvert_{(-1, -1, 4)}=\frac{\sqrt{yz}}{2\sqrt{x}}\bigg\rvert_{(-1, -1, 4)}=\frac{\sqrt{(-1)(4)}}{2\sqrt{-1}}=\frac{i\sqrt{4}}{2i}=\frac{2i}{2i}=1 $$
However, the official solution used the chain rule and arrived at a different result $$ \frac{\partial f}{\partial x}=\frac{\partial f}{\partial (xyz)}\cdot\frac{\partial (xyz)}{\partial x}=\frac{yz}{2\sqrt{xyz}} $$
$$ \frac{\partial f}{\partial x}\bigg\rvert_{(-1, -1, 4)}=\frac{yz}{2\sqrt{xyz}}\bigg\rvert_{(-1, -1, 4)}=\frac{(-1)(4)}{2\sqrt{(-1)(-1)(4)}}=\frac{-4}{2\sqrt{4}}=\frac{-4}{4}=-1 $$ Where was my mistake? If it was with the differentiation, what are some situations where I'm not allowed to take the power rule? If it was with complex numbers, what would be a general rule to prevent myself from making this mistake again? (I know $\sqrt{(-1)(-1)} \neq\sqrt{-1}\cdot\sqrt{-1}$ but that's about it)
The error has already been mentioned in a comment, but here are further details:
You are supposed to evaluate $\frac{\partial f}{\partial x}$ at a point where $x < 0$ and $yz < 0$. Therefore $\sqrt{xyz}$ is the square root of a positive number, but $\sqrt{yz}$ and $\sqrt{x}$ are square roots of negative numbers, and therefore $\sqrt{yz} \sqrt{x} \neq \sqrt{xyz}$; it is exactly the same thing as $\sqrt{-1}\sqrt{-1} \neq \sqrt{1}$, just with some positive factors inserted in all the square roots.