0

I have no idea if the question is going to be duplicated or not; what is the differential of the second order? If $\varphi(x,y)$ is a nice function that can be integrated and differentiated a desired number of times, then what is $d^2 \varphi$? Basically, it is explained to be the differential of the differential, so we apply recursion to the equality $$d\varphi = \frac{\partial \varphi}{\partial x}dx+\frac{\partial \varphi}{\partial y}dy$$ That is, $$d^2 \varphi =\frac{\partial}{\partial x}\left(\frac{\partial \varphi}{\partial x}dx+\frac{\partial \varphi}{\partial y}dy\right)dx+\frac{\partial}{\partial y}\left(\frac{\partial \varphi}{\partial x}dx+\frac{\partial \varphi}{\partial y}dy\right)dy$$ As you can see, I used the first formula for the expression on the right side of it. Following the same idea, we can acquire the expression for $d^3\varphi$. Wikipedia says that the exact differential can be defined as a map?? But my question is not related to it but related to why in books, the differentials are of the general form: $$d^n \varphi =\left(\frac{\partial \varphi}{\partial x}dx+\frac{\partial \varphi}{\partial y}dy\right)^n$$ I would like to apologize for such an unsuitable question but I do wish to discover the tiny truth about these expressions. We deal with TOTAL DIFFERENTIAL, not with exact - I mixed concepts...

sergei ivanov
  • 194
  • 2
    The extension you probably want here is the exterior derivative, which has $d^2 = 0$. In higher dimensions, the derivative is a matrix (or at least a vector), rather than a real-valued function, and so taking iterated derivatives doesn't quite make sense. You can talk about the matrix of second-order partial derivatives (the Hessian of a function), and the higher-order partial derivatives exist and satisfy a version of Taylor's theorem under mild restrictions on $\varphi$, but it's hard to assign a proper meaning to the last equation in your post as written. – anomaly Jun 10 '21 at 11:57
  • I would like to close the question and commence investigation but if someone having more experience wishes to help, I leave it for a while. I would not like to spend your time in vain. Thank you! – sergei ivanov Jun 10 '21 at 14:03

0 Answers0