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I was wondering what I'm doing wrong by considering $g(x,y)=x^2$ and then considering $\frac{\partial }{\partial x} \left(\frac{\partial g}{\partial (x^2)}\right) = \frac{\partial}{\partial x} (1) = 0 \neq \frac{\partial}{\partial (x^2)} \left( \frac{\partial g}{\partial x}\right) = \frac{1}{2x}$?

($\frac{\partial}{\partial (x^2)}$ denotes differentiation with respect to $x^2$)

A quarky name
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1 Answers1

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This is not the correct way to use partial derivatives in the first place. Partial derivatives are used when the two variable with respect to which you are differentiating are independent of each other.

Here, $x^2$ and $x$ are clearly not independent of each other, so Clairaut's theorem that $f_{xy} = f_{yx}$ does not hold.

$\textbf{Edit}$:

This is in response to the comment by the OP.

The LHS of the equation, $\frac{d}{dx} \frac{ d(g(x)) }{ dx^2 }$ is correct, it is clearly $0$ as $\frac{ d(g(x)) }{ dx^2 } = 1$ which is a constant and so further differentiation would give $0$. Note that I'm using $d$, not $\partial$. The only independent variable in this is $x$, the variable $x^2$ is not an independent variable and $y$ is not used anywhere. So your question is now just an application of the single-variable chain rule.

The RHS on evaluation gives, \begin{align} \frac{ d }{ d(x^2) } \frac{ d(g(x)) }{ dx } &= \frac{ d(2x) }{ d(x^2) } \\ &= \frac{ 2 (dx) }{ 2x (dx)} \\ &= \frac{1}{x} \end{align} Here, I've taken advantage of the Liebniz notation in differentiating the numerator and denominator separately, and quote-unquote "cancelling" the $dx$ as in a fraction.

So in reality, this is not a question of multivariable calculus, but single variable only. So these derivatives are clearly not equal, but they can be evaluated using the chain rule like usual.

Ash Ketchum
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  • Hi, thanks for the answer. That's a good point - makes sense. Would you still say that each of the two sides of the inequality above are correct, it's just that we shouldn't expect the partial derivatives to commute? – A quarky name Jul 31 '21 at 12:30
  • Thanks a lot, your answers have been super helpful. There was one last point I'd love to clear up. Essentially my questions came up when considering this function(al) $F[g(x_1,x_2,...,x_n),x_1,x_2,...,x_n, y,z]]$ (I guess you could really forget about listing $g$ as an argument of $F$ but I just wanted to stress that the $x_j$ appear in a combination which equals $g$. I wanted to find the mixed derivative $\frac{\partial }{\partial x_i } \left( \frac{\partial F[g(x_1,x_2,...,x_n),x_1,x_2,...,x_n, y,z]}{\partial g }\right)$ (I'm not sure the use of partials here is technically correct). – A quarky name Jul 31 '21 at 19:46
  • I was wondering (A) is this a valid thing to do and (B) if it is how would you differentiate the "loose" $x_j$ terms with respect ot $g$? Something like $\frac{\partial }{\partial g} = \sum \limits_{i}^{} \frac{\partial x_i}{\partial g} \frac{\partial }{\partial x_i}$ doesn't seem to give a reasonable result? I'm guessing $\frac{\partial x_i}{\partial g}$ doesn't make sense now that $g$ isn't a function of a single variable? – A quarky name Jul 31 '21 at 19:46