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I am wanting to take the derivative with respect to $x$ of $f(xy, x^2+3y-2)$, so, I use the chain rule to find $$ \frac{\partial}{\partial x} f(xy, x^2-2) = y \frac{\partial}{\partial xy}f(xy, x^2-2) + 2x\frac{\partial}{\partial (x^2-2)}f(xy, x^2-2) $$ With which I would then continue $$ \frac{\partial}{\partial xy} f(xy, x^2-2) = \frac{1}{x} \frac{\partial}{\partial y} f(xy, x^2-2) $$ However, when differentiating w.r.t $xy$, I am not communicating to the reader that $x^2-2$ is expected to remain a constant, and it would seem to them that I should rewrite $x^2-2$ as $\left(\frac{xy}{y}\right)^2-2$.

How do I use mathematical notation to communicate to another mathematician that $x^2-2$ remains constant when differentiating with respect to $xy$?

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You start out wanting $\frac{\partial}{\partial x}f(xy,x^2-2)$.

There is this function $f$ of two variables that I might call $u$ and $v$: $f=f(u,v)$. Here, $u$ is composed with $xy$ and $v$ with $x^2-2$. The chain rule says:

$$\begin{align} \frac{\partial}{\partial x}f(xy,x^2-2) &=\frac{\partial f}{\partial u}\cdot\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\cdot\frac{\partial v}{\partial x}\\ &=\frac{\partial f}{\partial u}\cdot y+\frac{\partial f}{\partial v}\cdot2x\\ \end{align}$$

And that's where I'd leave it until you know more about the nature of $f(u,v)$. You have $u$ and $v$, independent variables. No one will think they have a relation. Later, if you have some formula for $f(u,v)$ and you can actually write a formula for $\frac{\partial f}{\partial u}$ and $\frac{\partial f}{\partial v}$, you could substitute away the $u$s and $v$s with $xy$ and $x^2-2$. Or you could write it as

$$\left.\left(y\frac{\partial f}{\partial u}+2x\frac{\partial f}{\partial v}\right)\right\rvert_{u=xy,v=x^2-2}$$

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Let $\alpha(x,y)=xy$ and $\beta(x,y)=x^2-2$. Then, the function you're actually looking at is the composition $F(x,y)=f(\alpha(x,y),\beta(x,y))$. The chain rule tells you \begin{align} (\partial_1F)_{(x,y)} &=(\partial_1f)_{(\alpha(x,y),\beta(x,y))}\cdot (\partial_1\alpha)_{(x,y)} + (\partial_2f)_{(\alpha(x,y),\beta(x,y))}\cdot (\partial_1\beta)_{(x,y)}\\ &= (\partial_1f)_{(xy-x^2-2)}\cdot y + (\partial_2f)_{(xy-x^2-2)}\cdot 2x. \end{align} This is the strictest way of writing down the chain rule, and also the least ambiguous: the subscripts $1,2$ tell you along which direction you're calculating the partial derivatives, and the subscripts $(xy,x^2-2)$ etc tell you the point of evaluation of the derivatives. So, if you want to be 100% precise, this is how you could write the chain rule calculation.


If you want a slight mix with the Leibniz notation, you could also write this:

\begin{align} \frac{\partial}{\partial x} f(xy,x^2-2) &= (\partial_1f)_{(xy,x^2-2)}\cdot \frac{\partial (xy)}{\partial x} + (\partial_2f)_{(xy,x^2-2)}\cdot \frac{\partial (x^2-2)}{\partial x}\\ &=(\partial_1f)_{(xy-x^2-2)}\cdot y + (\partial_2f)_{(xy-x^2-2)}\cdot 2x. \end{align}

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