# Higher order derivatives and the chain rule

So here I have an assignment about higher order derivatives and the chain rule, and a relation to be proved: Show that for a rotation in the plane$$\begin{bmatrix}u\\v \end{bmatrix} =\begin{bmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{bmatrix} \begin{bmatrix}x\\y \end{bmatrix}$$

and any twice differentiable function $$f,$$ there holds $$\frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2} = \frac{\partial^2f}{\partial u^2}+\frac{\partial^2f}{\partial v^2}.$$ What I got so far is that $$f(u,v)=f(x\cos \theta-y\sin\theta, x\sin\theta + y\cos\theta)$$ from the matrix multiplication. But I don't really understand how to get $$\frac{\partial f}{\partial u}$$and $$\frac{\partial f}{\partial v}$$.

• I think you meant $(u,v)=(x\cos\theta-y\sin\theta,x\sin\theta+y\cos\theta)$, not $f(u,v)$, and $\dfrac{\partial f}{\partial u}=\dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial u}$ May 24 '20 at 21:55
• @J.W.Tanner Yes, I meant (u,v).. But what is $\frac{\partial x}{\partial u}$ and $\frac{\partial y}{\partial u}$ etc. since x and y do not consist u and v? Or did I misunderstand? May 24 '20 at 22:31
• you could invert the matrix to get $(x,y)=(u\cos\theta+v\sin\theta,-u\sin\theta+v\cos\theta)$ May 24 '20 at 22:37

The relation between $$(u,v)$$ and $$(x,y)$$ is written down by you, and then by the chain rule we could get
$$\frac {\partial f}{\partial x} = \frac {\partial f}{\partial u} \frac {\partial u}{\partial x}+ \frac {\partial f}{\partial v} \frac {\partial v}{\partial x} = \frac {\partial f}{\partial u}\cos\theta + \frac {\partial f}{\partial v}\sin\theta,$$ and likewise you can get $$\frac {\partial f}{\partial y}= -\frac {\partial f}{\partial u}\sin\theta+ \frac {\partial f}{\partial v}\cos\theta.$$ Following this and we can do chain rule again to obtain $$\frac {\partial^2f}{\partial x^2} = \frac {\partial^2f}{\partial u^2}\cos^2\theta+2 \frac {\partial^2 f}{\partial u\partial v}\cos\theta\sin\theta+ \frac {\partial^2f}{\partial v^2}\sin^2\theta$$ and $$\frac {\partial^2 f}{\partial y^2}= \frac {\partial^2f}{\partial u^2}\sin^2\theta-2 \frac {\partial^2 f}{\partial u\partial v}\cos\theta\sin\theta+ \frac {\partial^2f}{\partial v^2}\cos^2\theta,$$ by which the conclusion follows. Note that this proves the Laplacian is invariant under the rotation.