second-order total differential

Question

Let $f$ be a twice-differentiable function. Find the second-order total differential of the function $\phi(x, y, z) = f(u)$ if $u = xyz$. (Please use $f'(u), f''(u)$ signs.)

So what I was trying to do:

$$\partial u/\partial x = yz,\; \partial u/\partial y = xz,\; \partial u/\partial z = xy$$

Then

$$\partial \phi/\partial x = \partial f/\partial u \, \partial u/\partial x = f'(u) \, yz\\ \partial \phi/\partial y = \partial f/\partial u \, \partial u/\partial y = f'(u) \, xz\\ \partial \phi/\partial z = \partial f/\partial u \, \partial u/\partial z = f'(u) \, xy$$

Now, i need to find the second-order partial derivatives of $\phi$ with respect to $x$, $y$, and $z$. Here's the problem: i don't understand how to do it.. it's $\partial^2\phi/\partial x^2 = (\partial/\partial x)(\partial \phi/\partial x)$. what to do next? i can't understand how to find them and get to the answer. Also is it true that $$d^2\phi = (\partial^2\phi/\partial x^2) \, (dx)^2 + (\partial^2\phi/\partial y^2) \, (dy)^2 + (\partial^2\phi/\partial z^2) \, (dz)^2 + 2(\partial^2\phi/\partial x\partial y) \, dx \, dy + 2(\partial^2\phi/\partial x\partial z) \, dx \, dz + 2(\partial^2\phi/\partial y\partial z) \, dy \, dz \text?$$ I get the formula but why are $(dx)^2$ squared?

Answer 1

You saw that to differentiate $ \phi ( x , y , z ) $ (which is the same as $ f ( u ) $) with respect to $ x $, you would first differentiate it with respect to $ u $, then multiply this by $ \partial u / \partial x $. You can do a similar thing when it comes to differentiating $ \partial \phi / \partial x $ with respect to $ x $. At least, that's what you do when you get to the part involving $ u $; for the part involving $ y $ and $ z $, you can just treat them as constants (since you're differentiating with respect to $ x $). (And if there was a part involving $ x $ in $ \partial \phi / \partial x $, then you would use $ \partial x / \partial x = 1 $ there, but there isn't a part like that.) So $$ \partial ( \partial \phi / \partial x ) / \partial x = \frac { \partial f ' ( u ) y z } { \partial x } = y z \frac { \partial f ' ( u ) } { \partial x } = y z \frac { d f ' ( u ) } { d u } \frac { \partial u } { \partial x } = y z f ' ' ( u ) y z = f ' ' ( u ) y ^ 2 z ^ 2 \text . $$ And similarly for the other eight partial derivatives. (Well, the other five independent ones; but if you find all nine of them, then you can double-check that the mixed partials are equal as they should be.)

As for the reason for the formula for the second differential, the comment by @eyeballfrog is correct, although to get the formula that you wrote, you also have to assume that the second derivatives of $ x $, $ y $, and $ z $ with respect to $ t $ are all zero. If you don't do that, then you get the total total second differential $$ d ^ 2 \phi = \frac { \partial ^ 2 \phi } { \partial x ^ 2 } \, d x ^ 2 + \frac { \partial ^ 2 \phi } { \partial y ^ 2 } \, d y ^ 2 + \frac { \partial ^ 2 \phi } { \partial z ^ 2 } \, d z ^ 2 + 2 \frac { \partial ^ 2 \phi } { \partial x \partial y } \, d x \, d y + 2 \frac { \partial ^ 2 \phi } { \partial x \partial z } \, d x \, d z + 2 \frac { \partial ^ 2 \phi } { \partial y \partial z } \, d y \, d z + \frac { \partial \phi } { \partial x } \, d ^ 2 x + \frac { \partial \phi } { \partial y } \, d ^ 2 y + \frac { \partial \phi } { \partial z } \, d ^ 2 z \text . $$ If you know that $ x $, $ y $, and $ z $ are the independent variables that you're going to use at bottom, then you can safely set their higher differentials to zero and so ignore the last three terms; but if there's any chance that you might make them functions of something else (for real, not just to derive the formula as @eyeballfrog said), then you can't leave those terms out.

Although a quick way to answer why $ d x $ needs to be squared (in the first term of $ d ^ 2 \phi $) is so that the units match, to cancel the fact that $ \partial x $ is squared in the denominator of $ \partial ^ 2 \phi / \partial x ^ 2 $. (In fact, the terms of the total version of $ d ^ 2 \phi $ are all the possible ways to write a term in which the units of $ x $, $ y $, and $ z $ all cancel, leaving the same units as $ \phi $, and in which the total infinitesimal order is $ 2 $.)

Answer 2

Answer:

It can be done just like so:

$$d^2u=d(du)=d\sum_i\partial_iu\ dx^i=\sum_i\partial_iu\ d^2x^i+d(\partial_iu)\ dx^i\\=\sum_{i}\left(\partial_iu\ d^2x^i+\sum_j \partial_j\partial_iu\ dx^jdx^i\right).$$ Now, to obtain your wanted result: $$d^2f(xyz)=d(df(xyz))=d(f'(xyz)(yzdx+xzdy+xydz))\\ = f''(xyz)(yzdx+xzdy+xydz)^2+f'(xyz)d(yzdx+xzdy+xydz)\\ = \boxed{\boxed{(yz)^2f''(xyz)dx^2+(xz)^2f''(xyz)dy^2+(xy)^2f''(xyz)dz^2\\+2((xyz^2f''(xyz)+zf'(xyz))dxdy\\+(yzx^2f''(yzx)+xf'(yzx))dydz\\+(yzx^2f''(xyz)+yf'(xyz))dzdx)}}.$$