I am revising for my first year Calculus examination. The following question is on a past paper and I am given the solution, however I am struggling to make sense of it:
Let $V(x,y)$ be a differentiable function and set $$W(r,\varphi):=V(r\cos\varphi,r\sin\varphi).$$ Apply the chain rule to show that $$ \left(\frac{\partial V}{\partial x}\right)^{\!2}+ \left(\frac{\partial V}{\partial y}\right)^{\!2}= \left(\frac{\partial W}{\partial r}\right)^{\!2}+ \frac{1}{r^2}\left(\frac{\partial W}{\partial \varphi}\right)^{\!2} $$
I am told that $V_r=V_x*x_r+V_y*y_r$ but do not see why.
The multivariable chain rule tells us that \begin{equation} \frac{\partial W(r,\phi)}{\partial r} = \frac{\partial V(r \cos \phi,r \sin \phi)}{\partial x} \cos \phi + \frac{\partial V(r \cos \phi, r \sin \phi)}{\partial y} \sin \phi \end{equation} and \begin{equation} \frac{\partial W (r,\phi)}{\partial \phi} = \frac{\partial V(r \cos \phi,r \sin \phi)}{\partial x} (-r \sin \phi) + \frac{\partial V(r \cos \phi, r \sin \phi)}{\partial y} (r \cos \phi). \end{equation}
Using these equations, it's easy to check that the expression \begin{equation} \left(\frac{\partial W(r,\phi)}{\partial r} \right)^2 + \frac{1}{r^2} \left(\frac{\partial W(r,\phi)}{\partial \phi} \right)^2 \end{equation} simplifies to \begin{equation} \left( \frac{\partial V(r \cos \phi, r \sin \phi)}{\partial x} \right)^2 + \left(\frac{\partial V(r \cos \phi, r \sin \phi)}{\partial y} \right)^2. \end{equation}