For this Question is if were to do $U_r$ after the first partial derivative. It will result in a function $U_r(x,y,θ)$ why is it that upon the second time I do a partial derivative, I need not consider it as the following $U_{rr}=\frac{∂Ur}{∂x}·\frac{∂x}{∂r}+\frac{∂Ur}{∂y}·\frac{∂y}{∂r}+\frac{∂Ur}{∂θ}·\frac{∂θ}{∂r}$? Instead what was considered to be the answer was $U_{rr}=\frac{∂Ur}{∂x}·\frac{∂x}{∂r}+\frac{∂Ur}{∂y}·\frac{∂y}{∂r}.$ Do we not need to consider the last term? and if so how would I set up a function to do the partial derivative of $\frac{∂θ}{∂r}$? Thanks
The function $u$ is constructed as a composition $$ u=f(x,y),\qquad x=r\cos\theta,\ y=r\sin\theta. $$ If we write all the variable dependencies carefully we will get $$ u(r,\theta)=f(x(r,\theta),y(r,\theta)). $$ Now using the chain rule $$ \frac{\partial u}{\partial r}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}=\frac{\partial f}{\partial x}\cos\theta+\frac{\partial f}{\partial y}\sin\theta. $$ Here the partial derivatives of $f$ are again compositions that depend on $x(r,\theta)$ and $y(r,\theta)$, which makes the second differentiation slightly more messy, but it is the same chain rule $$ \frac{\partial^2 u}{\partial r^2}=\frac{\partial}{\partial r}\left(\frac{\partial f}{\partial x}\right)\cos\theta+\frac{\partial}{\partial r}\left(\frac{\partial f}{\partial y}\right)\sin\theta $$ where e.g. $$ \frac{\partial}{\partial r}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)\frac{\partial x}{\partial r}+\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)\frac{\partial y}{\partial r} $$ etc