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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?

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  • $\begingroup$ Let $\phi(x,y) = {\partial f(x,y) \over \partial y}$, then ${d \phi \over dx} = {\partial \phi(x,y) \over \partial x} + {\partial \phi(x,y) \over \partial y} {dy \over dx}$. $\endgroup$ – copper.hat Dec 29 '13 at 5:34
  • $\begingroup$ @copper.hat, nice, but not necessarily intuitive. $\endgroup$ – cygorx Dec 29 '13 at 5:35
  • $\begingroup$ @cygorx: I'm trying :-). I guess the point is that both partials of $f$ are still functions of both $x$ and $y$. $\endgroup$ – copper.hat Dec 29 '13 at 5:35

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