# Using the chain rule to find second order partial derivatives

Given the application $$u(r,\phi):=v(r\cos\phi, r\sin \phi)$$ I need to find by the chain rule an expression for $$\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2}$$ in terms of second partial derivatives of $$u.$$ I found first the Jacobian matrix based on the transformation function $$(r,\phi)\rightarrow (x,y)=(r\cos \phi, r \sin \phi).$$ Since this application is not a function, I do not know if one has to go through the Hessian matrix to find the second partial derivatives. Can you provide me some support or a solution proposal ? Thanks.

• Where do $x,y$ feature in all this? Commented Sep 20, 2022 at 23:53
• $(x,y)$ is represented by polar coordinates. I need to thus represent the two second order partial derivatives in terms of partial derivatives of $u$ and the polar coordinates. Commented Sep 21, 2022 at 6:40
• You will confuse yourself and everyone else unless you write something sensible like "$u(r,\phi)=v(x,y)$ where $x=r\cos\phi$, $y=r\sin\phi$". The result is in every book: what is the problem? Commented Sep 21, 2022 at 6:43
• @user996159 Have you made any progress? Commented Oct 5, 2022 at 3:31

Hint: Since you want an expression of the second order partial derivatives of $$u$$, begin by finding what they are in terms of the p.d. of $$v$$ and the derivatives of $$x$$ and $$y$$.
Reason: the first and second order partial derivatives of $$x, y$$ with respect to $$r, \phi$$ are much easier to find than the inverse, and trigonometric identities may be useable.$$\dfrac{\partial x}{\partial r}=\cos\phi~, \dfrac{\partial x}{\partial \phi}=-r\sin\phi~, \dfrac{\partial^2 x}{\partial r~^2}=0~,\textit{et cetera}\ldots$$