The question asks us to prove that, if $ u = f(x, y) $, being $ x = e^scos(t) $ and $ y = e^ssin(t) $, show that $$ (\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2 = e^{-2s}((\frac{\partial u}{\partial s})^2+(\frac{\partial u}{\partial t})^2) $$
I know that about the chain rule with $ \frac{\partial u}{\partial t} $ and applying the derivative in terms of t, but what about $ \frac{\partial u}{\partial x} $ when no function is given? How would you go about solving this?
We start from the right hand side, $$ \begin{align} \frac{\partial u}{\partial s} &= \frac{\partial u}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial s} \\ &= u_x e^s \cos t + u_y e^s \sin t \end{align} $$ and \begin{align} \frac{\partial u}{\partial t} &= \frac{\partial u}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial t} \\ &= -u_x e^s \sin t + u_y e^s \cos t \end{align} Hence, $$ e^{-2s} \left( (u_t)^2 + (u_s)^2 \right) = (u_x)^2 + (u_y)^2 $$