1
$\begingroup$

How does one rigorously define partial derivatives? If for $f: \mathbb{R}^n \to \mathbb{R}$ we define $$\frac{\partial^rf}{\partial x_1^{\alpha_1} \dots \partial x_n^{\alpha_n}},$$ where $\alpha_1 + \dots + \alpha_n = r$, as a result of differentiating $\alpha_n$ times w.r.t. $x_n$, $\alpha_{n-1}$ times w.r.t. $x_{n-1}$ and so on, how can we prove that this is the same as applying rth total derivative to corresponding basis vectors? I want something like $$(D^rf)_p(e_{i_1})\dots(e_{i_r}) = \left(\frac{\partial^rf}{\partial x_1^{\alpha_1} \dots \partial x_r^{\alpha_r}}\right)_p$$ to be true.

$\endgroup$
  • $\begingroup$ What is this $\partial^n$ in the numerator? $\endgroup$ – zhw. Jul 2 '17 at 7:40
  • $\begingroup$ @zhw, sorry, bad notation. Edited. $\endgroup$ – edubrovskiy Jul 2 '17 at 8:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.