How do differentiate a partial derivative

Question

I have $f(x, y)$ where $y$ is a function of $x$. So $f(x,y)=f(x, y(x))$. How do I differentiate the partial derivative $$\frac{d}{dx}\left(\frac{\partial f}{\partial y}\right)=?$$ $$\frac{d}{dy}\left(\frac{\partial f}{\partial x}\right)=?$$

Answer 1

When you find yourself getting confused in a situation like this, it is time to get explicit and pedantic.

Write $f$ as $f(X,Y)$, where
$ X(x) = x$
$ Y(x) = y(x)$.

Now you can consider the function $g(x) = f(X(x), Y(x))$, if you want to. But you should write anything you're curious about in terms of these variables, to make sure that what you're asking about is meaningful.

Notice that if you do that, the only meaningful lowest-level derivatives are

$$\frac{\partial f}{\partial X}, \frac{\partial f}{\partial Y}, \frac{dY}{dx} \textrm{ [or } y'(x)\textrm{], and }\frac{dX}{dx} \textrm{ [or } 1\textrm{]}$$ along with their higher-order relatives. I'm still thinking about it, but I'm starting to suspect that the two things you've written aren't actually meaningful - $\frac{d}{dy}\left(\frac{\partial f}{\partial x}\right)$ looks particularly suspicious. You should provide more detail on where they came from.

Added after some comments:

Okay, it sounds like you're curious about the function $\frac{\partial f}{\partial Y}$, which I'm going to call

$$h(X,Y) = \frac{\partial f}{\partial Y}(X,Y)$$

And you say you want to treat $X$ and $Y$ as dependent variables, which means you want to just consider them along the path in $(X,Y)$ space defined by $(x,y(x))$. So the function you want to differentiate is

$$k(x) = h(X(x), Y(x))$$

(I'll note that you say you want $dk$, where "$d$" is usually some sort of total derivative. But when your function is a function of only one variable, say $s(t)$, then $ds$ is just $s'(t)dt$, i.e. it's not that much different from just computing $\frac{d}{dt}$ of $s$.)

So, back to $k$. Looking at its definition, we can see that it is set up perfectly for the chain rule:

$$\frac{d}{dx}k = \frac{\partial h}{\partial X}\frac{dX}{dx} + \frac{\partial h}{\partial Y}\frac{dY}{dx}$$

and substituting back all the intermediary function names we used to help us keep things straight, we get

$$\frac{d}{dx}k = \frac{\partial^2 f}{\partial X \partial Y}\cdot 1 + \frac{\partial^2 f}{\partial Y^2}\cdot y'(x)$$

We can, at this point, go back and look at the two expressions you originally asked about:

• The first, $\frac{d}{dx}\left(\frac{\partial f}{\partial y}\right)$, roughly corresponds to the first term of our answer, although we see that "$\partial x$" is what makes sense, not "$dx$", and also that it really isn't "equal" to something else - that's as simple an expression we can have without knowing more about $f$.

• The second, $\frac{d}{dy}\left(\frac{\partial f}{\partial x}\right)$, doesn't seem to have been involved at all. Which is lucky for me, because I still can't think what it might represent.