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In my textbook there are two questions on finding second order derivatives for a function $f(x(u, v), y(u, v)$.

In the first question I have to take:

$ (\frac{\partial}{\partial x})(y\frac{\partial f}{\partial x}) $

$ = \frac{\partial f}{\partial x} \frac{\partial y}{\partial y} + y \frac{\partial^2f}{\partial x^2} $ (applying the chain rule)

$ = \frac{\partial f}{\partial x} + y \frac{\partial^2f}{\partial x^2} $

Then in the second question I have to take:

$ (\frac{\partial}{\partial x})(\frac{\partial f}{\partial x}u) $

$ = u\frac{\partial^2f}{\partial x^2} $ (not applying the chain rule)

I do not understand why the chain rule must be applied in the first case but not the second, and there is no explanation in my textbook. I think it's to do with the first case being differentiating wrt the a variable from the same coordinate system, and the second being from different coordinate systems, but I can't get any further that that. Any help would be greatly appreciated!

Thank you!

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2 Answers 2

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We are considering $x, y$ to be independent variables, and $u, v$ to be independent, but both $x,y$ as functions of $u, v$. So ... firstly we are not using the Chain Rule:

$\begin{align} &~\dfrac{\partial~~}{\partial x}\left(y\cdot\dfrac{\partial f}{\partial x}\right)\\ =~&~ \dfrac{\partial y}{\partial x}\cdotp\!\dfrac{\partial f}{\partial x} + y\cdot \dfrac{\partial^2 f}{\partial x~^2}&&:\textbf{Product Rule}\\=~&~0+y\cdot \dfrac{\partial^2 f}{\partial x~^2}&&:\text{via variable independence}\end{align}$

But for the second, unless $x$ and $u$ are independent then the additional term will not vanish. Since $x$ is given as a function of $u$ and $v$, this may well not be the case. In such case we need the Jacobian Inverse Function Theorem.

$\begin{align} &~\dfrac{\partial~~}{\partial x}\left(\dfrac{\partial f}{\partial x}\cdot u\right)\\ =~&~ \dfrac{\partial^2 f(x,y)}{\partial x~^2}\cdot u+\dfrac{\partial f(x,y)}{\partial x}\cdot\dfrac{\partial u}{\partial x} &&:\text{Product Rule}\\=~&~\dfrac{\partial^2 f(x,y)}{\partial x~^2}\cdot u+\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial y}{\partial v}\div\left(\dfrac{\partial x}{\partial u}\dfrac{\partial y}{\partial v}-\dfrac{\partial x}{\partial v}\dfrac{\partial y}{\partial u}\right)&&:\textbf{Inverse Function Theorem}\end{align}$

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Since $u$ is an independent variable, therefore $$\frac{\partial u}{\partial x}=0.$$ Therefore $$\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}u\right)=u\frac{\partial f}{\partial x^2}.$$

However, I do think you have made a mistake in typing the first question.

$$ \frac{\partial}{\partial x}\left(y\frac{\partial f}{\partial x}\right) \neq \frac{\partial f}{\partial x} \frac{\partial y}{\partial y} + y \frac{\partial^2f}{\partial x^2} $$

How did you get $\dfrac{\partial y}{\partial y}?$

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  • $\begingroup$ Thank you so much for your answer! Is this the same logic as for $\frac{\partial x}{\partial y} = 0$? So really you are applying the chain rule even in the second case, it is just that the $\frac{\partial u}{\partial x}$ term vanishes? Yes, I'm sorry! I typed in the wrong terms for the first question, thank you for pointing it out :) $\endgroup$ Commented Apr 23, 2020 at 21:32
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    $\begingroup$ Yes, you're right. $x$ depends on $u$ and $v$ but not on $y.$ So $\frac{\partial x}{\partial y}=0.$ $\endgroup$ Commented Apr 23, 2020 at 22:12
  • $\begingroup$ @mathopeas Because $x$ depends on $u$, therefore $u$ is not independent of $x$. To find $\dfrac{\partial u}{\partial x}$ use the inversion theorem:$$\left[\dfrac{\partial\langle u, v\rangle}{\partial\langle x, y\rangle}\right] = \left[\dfrac{\partial\langle x, y\rangle}{\partial\langle u, v\rangle}\right]^{-1}$$ $\endgroup$ Commented Apr 23, 2023 at 8:48

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