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This is the total differential $$df=dx\frac {\partial f}{\partial x}+dy\frac {\partial f}{\partial y}.$$

How do we take higher orders of total differential, $d^2 f=$?

Suppose I have $f(x,y)$ and I want the second order total differential $d^2f$?

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  • $\begingroup$ Product and chain rule just how you did to get the first one. Now imagine that $df$ is a function $g(x,y)$, which is formed by the sum of two products, and get the differential of $g$. $\endgroup$ – MyUserIsThis Jun 30 '13 at 14:03
  • $\begingroup$ Like $d^2f=dx\wedge(\frac{\partial^2f}{\partial x\partial x}dx+\frac{\partial^2f}{\partial x\partial y}dy)+dy\wedge(\frac{\partial^2f}{\partial y\partial x}dx+\frac{\partial^2f}{\partial y\partial y}dy)=dx\wedge dy(\frac{\partial^2f}{\partial x\partial y}-\frac{\partial^2f}{\partial y\partial x})$? $\endgroup$ – awllower Jun 30 '13 at 14:05
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$d=dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}$ is a differential operator that when applied to $f$ gives $df=dx\frac{\partial f}{\partial x}+dy\frac{\partial f}{\partial y}$

$d^2=\left(dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}\right)\left(dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}\right)$

$d^2f=\left(dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}\right)\left(dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}\right)f=\left(dx^2\frac{\partial^2}{\partial x^2}+dy^2\frac{\partial^2}{\partial y^2}+2dx dy\frac{\partial^2}{\partial x\partial y}\right)f=dx^2\frac{\partial^2 f}{\partial x^2}+dy^2\frac{\partial^2 f}{\partial y^2}+2dx dy\frac{\partial^2 f}{\partial x\partial y}$

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  • $\begingroup$ Do you mean, when $f$ is $C^2$, $d^2f=dx^2\frac{\partial^2 f}{\partial x^2}+dy^2\frac{\partial^2 f}{\partial y^2}+2dx dy\frac{\partial^2 f}{\partial x\partial y}$? Even if so, you must have forgotten that $dxdy+dydx=0$. Regards. $\endgroup$ – awllower Jul 1 '13 at 2:41
  • $\begingroup$ @awllower You mean $dx dy\wedge dy dx=0$ not $dx dy+dy dx=0$ $\endgroup$ – llllllllllllllllllllllllllllll Jul 1 '13 at 7:28
  • $\begingroup$ No, I mean $dx\wedge dy+dy\wedge dx=0$, as the wedge product is anti-symmetric. $\endgroup$ – awllower Jul 1 '13 at 12:28
  • $\begingroup$ Yes, but as you saw, I avoided the use of the wedge product because I think @dan only want to use elementary calculus. $\endgroup$ – llllllllllllllllllllllllllllll Jul 1 '13 at 16:34
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    $\begingroup$ Probably a basic question, but why is $d^2 f$ (ie $d(df)$) the same as multiplying $d$ with $d$ and then applying that result to $f$? Ie how does the composition of differential operators get reduced to a multiplication of those operators? $\endgroup$ – user434180 Nov 3 '16 at 0:55
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I will assume that you are referring to the Fréchet derivative. If $U\subseteq\mathbb{R}^n$ is an open set and we have functions $\omega_{j_1,\dots,j_p}:U\to\mathbb{R}$, then $$ D\left(\sum_{j_1,\dots,j_p} \omega_{j_1,\dots,j_p} dx_{j_1}\otimes\cdots\otimes dx_{j_p}\right) = \sum_{j_1,\dots,j_p}\sum_{j=1}^n \frac{\partial\omega_{j_1,\dots,j_p}}{\partial x_{j}} dx_j\otimes dx_{j_1}\otimes\cdots\otimes dx_{j_p}. $$

Here $dx_{i}$ is the projection onto the $i$th coordinate, and if $\alpha,\beta$ are multilinear forms then $\alpha\otimes\beta$ is the multilinear form defined by $(\alpha\otimes\beta)(x,y)=\alpha(x)\beta(y)$.

For example, let $f(x,y)=x^3+x^2 y^2+y^3$. Then \begin{align} Df&=(3x^2+2xy^2)dx+(2x^2y+3y^2)dy; \\ D^2f&=(6x+2y^2)dx\otimes dx+4xy(dx\otimes dy+dy\otimes dx)+(2x^2+6y)dy\otimes dy; \\ D^3f&=6dx\otimes dx\otimes dx+4y(dx\otimes dx\otimes dy+dx\otimes dy\otimes dx+dy\otimes dx\otimes dx)\\ &\qquad+4x(dx\otimes dy\otimes dy+dy\otimes dx\otimes dy+dy\otimes dy\otimes dx) \\ &\qquad+6dy\otimes dy\otimes dy. \end{align}

Since $D^p f(x)$ is always a symmetric multilinear map if $f$ is of class $C^p$, you might want to simplify the above by using the symmetric product (of tensors).

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  • $\begingroup$ This answer is good because you're doing the total derivative of the total derivative, rather than just the exterior derivative of the total derivative. $\endgroup$ – Toby Bartels yesterday
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Edit: I interpret your $d^2 f$ as $d^2f=d(df)$.

If $f$ is $C^2(\Omega)$, with $(x,y)\in\Omega$, then

$$d(df)=dy\wedge dx \frac{\partial^2 f}{\partial y\partial x}+dx\wedge dy \frac{\partial^2 f}{\partial x\partial y}=\left(\frac{\partial^2 f}{\partial x\partial y}- \frac{\partial^2 f}{\partial y\partial x} \right)dx\wedge dy=0,$$

as $dx \wedge dy=-dy\wedge dx$ and $\frac{\partial^2 f}{\partial x\partial y}= \frac{\partial^2 f}{\partial y\partial x}$ on $\Omega$.

If $f$ is not $C^2$ at $(x,y)$, then we arrive at

$$d(df)=\left(\frac{\partial^2 f}{\partial x\partial y}- \frac{\partial^2 f}{\partial y\partial x} \right)dx\wedge dy. $$

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  • $\begingroup$ But OP did not assume that $f\in C^2$, so that the result is not necessarily $0$, right? $\endgroup$ – awllower Jun 30 '13 at 14:22
  • $\begingroup$ @awlower, right. I used the Schwarz lemma. If $f$ is not $C^2$ then we have just the first equality. $\endgroup$ – Avitus Jun 30 '13 at 16:19
  • $\begingroup$ @awlower I add this comment in my answer. $\endgroup$ – Avitus Jun 30 '13 at 16:48

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