I'm struggling to understand which are the partial derivatives of the component functions of
$$F(x,y,z) = Pi + Qj + Rk$$
This question came up when I looking the solution of the following exercise:
Determine if the the vector field is conservative or not of the following function: $$F(x,y,z)=y^2z^2i + 2xyz^3j + 3xy^2z^2k$$
The theorem used to solve is:
If F is a vector field defined on all on $R^3$ whose component functions have continuous partial derivatives and curl F = 0, then F is a conservative vector field.
The final answer is yes (since $curl = 0$ and the space is $R^3$, as the book says), but it doesn't show the partial derivatives of $P$, $Q$ and $R$. What is the first derivative of $P$? And the second one?
In $$ F(x,y,z) = y^2z^2 i + 2 x y z^3 j + 3xy^2z^2 k \text{,} $$ we have (by finding the coefficients of $i$, $j$, and $k$), \begin{align*} P(x,y,z) &= y^2 z^2 \\ Q(x,y,z) &= 2 x y z^3 \\ R(x,y,z) &= 3xy^2z^2 \end{align*} Then, for example, \begin{align*} \frac{\partial P}{\partial x} &= 0 \\ \frac{\partial P}{\partial y} &= 2y z^2 \\ \frac{\partial P}{\partial z} &= 2y^2 z \text{.} \end{align*} For the $x$ partial derivative, recall that when we treat $x$ as the independent variable, we treat $y$ and $z$ as constants, so this is the derivative of a constant, hence zero. For the $y$ partial derivative, $y$ is the variable and $z$ is a constant, so the power rule only acts on $y^2$ and the $z^2$ passes through the constant multiple rule. For the $z$ partial derivative, $z$ is the variable and $y$ is a constant, so the power rule only acts on $z^2$ and the $y^2$ passes through the constant multiple rule.