Intuition behind second total derivative

Question

If we want to find the total derivative of a function $f(x, y)$, we can express it in terms of the function's partial derivatives as follows: $$\frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}$$ This formula fits very well with my intuition. The change in $f$ with respect to $x$ is the sum of the change due only to $x$ plus the change due to $y$ scaled by the relationship between $y$ and $x$. However, when we wish to take the second total derivative, the analogous formula is much less appealing: $$\frac{d^2f}{dx^2}=\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}\left(\frac{dy}{dx}\right)^2+\frac{\partial f}{\partial y}\frac{d^2y}{dx^2}+2\frac{\partial^2 f}{\partial x \partial y}\frac{dy}{dx}$$ Suddenly, two new terms have popped up, with a square of a derivative and a mixed partial derivative. Does there exist a nice intuitive explanation for the terms in this equation, in the same way an intuitive explanation for the first equation was presented above?

Answer 1

This should be clearer if you restore the symmetry between $ x $ and $ y $ by making them both functions of an independent variable $ t $. Then you'll have even more terms, but they'll fit a clearer pattern.

So the 1st total derivative is $$ \frac { \mathrm d f } { \mathrm d t } = \frac { \partial f } { \partial x } \frac { \mathrm d x } { \mathrm d t } + \frac { \partial f } { \partial y } \frac { \mathrm d y } { \mathrm d t } \text . $$ Hopefully, this makes sense; $ f $ can vary (relative to $ t $) either by varying relative to $ x $ (which then varies relative to $ t $) or by varying relative to $ y $ (which then varies relative to $ t $). It's also nice to look at some simple examples that focus attention on the individual terms: if $ f = 2 x $ and $ x = 3 t $, then $ f = 6 t $, and we have $ 6 = 2 \cdot 3 + 0 $; similarly, if $ f = 2 y $ and $ y = 3 t $, then $ f = 6 t $ again, but now we have $ 6 = 0 + 2 \cdot 3 $. Of course, if $ x = t $, then $ \mathrm d x / \mathrm d x = 1 $, so this reduces to $ \mathrm d f / \mathrm d x = \partial f / \partial x + \partial f / \partial y \cdot \mathrm d y / \mathrm d x $.

OK, now the 2nd total derivative is $$ \frac { \mathrm d ^ 2 f } { \mathrm d t ^ 2 } = \frac { \partial ^ 2 f } { \partial x ^ 2 } \bigg ( \frac { \mathrm d x } { \mathrm d t } \bigg ) ^ 2 + \frac { \partial ^ 2 f } { \partial y \, \partial x } \frac { \mathrm d y } { \mathrm d t } \frac { \mathrm d x } { \mathrm d t } + \frac { \partial f } { \partial x } \frac { \mathrm d ^ 2 x } { \mathrm d t ^ 2 } + \frac { \partial ^ 2 f } { \partial x \, \partial y } \frac { \mathrm d x } { \mathrm d t } \frac { \mathrm d y } { \mathrm d t } + \frac { \partial ^ 2 f } { \partial y ^ 2 } \bigg ( \frac { \mathrm d y } { \mathrm d t } \bigg ) ^ 2 + \frac { \partial f } { \partial y } \frac { \mathrm d ^ 2 y } { \mathrm d t ^ 2 } \text . $$

(You can prove this by calculating the derivative of $ \mathrm d f / \mathrm d t $ above, applying (in turn) the Sum Rule and the Product Rule (once in each term), and then applying the rule for the 1st total derivative itself when the time comes to differentiate the partial derivatives. But I'll assume that you're OK with the expression as it stands and just want to get an intuitive understanding of it.)

So, there are six ways that $ f $ can vary to second order (relative to $ t $): by varying to second order relative to $ x $ (which then varies relative to $ t $); by varying relative to $ x $ and then relative to $ y $ (each of which varies relative to $ t $); by varying relative to $ x $, which then varies to second order (relative to $ t $); by varying with respect to $ y $ and then $ x $ (each of which varies relative to $ t $); by varying to second order relative to $ y $ (which then varies relative to $ t $); and by varying relative to $ y $, which then varies to second order (relative to $ t $). For examples that highlight each term, consider these:

• $ f = 2 x ^ 2 $, $ x = 3 t $; then $ f = 1 8 t ^ 2 $, and we have $ 3 6 = 4 \cdot 3 ^ 2 + 0 + 0 + 0 + 0 + 0 $.
• $ f = 2 x y $, $ x = 3 t $, $ y = 4 t $; then $ f = 2 4 t ^ 2 $, and we have $ 4 8 = 0 + 2 \cdot 3 \cdot 4 + 0 + 2 \cdot 4 \cdot 3 + 0 + 0 $.
• $ f = 2 x $, $ x = 3 t ^ 2 $; then $ f = 6 t ^ 2 $, and we have $ 1 2 = 0 + 0 + 2 \cdot 6 + 0 + 0 + 0 $.
• (Repeat, see 2 lines above.)
• $ f = 2 y ^ 2 $, $ y = 3 t $; then $ f = 18 t ^ 2 $ again, but now we have $ 3 6 = 0 + 0 + 0 + 0 + 4 \cdot 3 ^ 2 + 0 $.
• $ f = 2 y $, $ y = 3 t ^ 2 $; then $ f = 6 t ^ 2 $ again, but now we have $ 1 2 = 0 + 0 + 0 + 0 + 0 + 2 \cdot 6 $.

Of course, by the symmetry of mixed partial derivatives, this simplifies to $$ \frac { \mathrm d ^ 2 f } { \mathrm d t ^ 2 } = \frac { \partial ^ 2 f } { \partial x ^ 2 } \bigg ( \frac { \mathrm d x } { \mathrm d t } \bigg ) ^ 2 + \frac { \partial f } { \partial x } \frac { \mathrm d ^ 2 x } { \mathrm d t ^ 2 } + 2 \frac { \partial ^ 2 f } { \partial x \, \partial y } \frac { \mathrm d x } { \mathrm d t } \frac { \mathrm d y } { \mathrm d t } + \frac { \partial ^ 2 f } { \partial y ^ 2 } \bigg ( \frac { \mathrm d y } { \mathrm d t } \bigg ) ^ 2 + \frac { \partial f } { \partial y } \frac { \mathrm d ^ 2 y } { \mathrm d t ^ 2 } \text , $$ but then the pattern is not quite as clear. Finally, if $ x = t $, then not only do some terms simplify because $ \mathrm d x / \mathrm d x = 1 $, but one term goes away entirely because $ \mathrm d ^ 2 x / \mathrm d x ^ 2 = 0 $, giving $$ \frac { \mathrm d ^ 2 f } { \mathrm d x ^ 2 } = \frac { \partial ^ 2 f } { \partial x ^ 2 } + 2 \frac { \partial ^ 2 f } { \partial x \, \partial y } \frac { \mathrm d y } { \mathrm d x } + \frac { \partial ^ 2 f } { \partial y ^ 2 } \bigg ( \frac { \mathrm d y } { \mathrm d x } \bigg ) ^ 2 + \frac { \partial f } { \partial y } \frac { \mathrm d ^ 2 y } { \mathrm d x ^ 2 } \text , $$ which is the formula that you were asking about. This is much shorter, but the pattern is largely hidden now.