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Let $z = f(x, y)$ be a function with continuous partial derivatives. Let $x = e^r\cos(\theta)$ and $y = e^r\sin(\theta)$. By finding all relevant partial derivatives and using the chain rule, show that $$\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2=e^{-2r}\left[\left(\frac{\partial z}{\partial r}\right)^2+\left(\frac{\partial z}{\partial \theta}\right)^2\right]$$ First I found the $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$, which I took to mean that I needed to to find $\frac{dz}{dr}+\frac{dr}{d\theta}$ because of the chain rule, so I got: $$\frac{dx}{dr}+\frac{dx}{d\theta}=e^r \cos(\theta)-e^r \sin(\theta)$$ $$\frac{dy}{dr}+\frac{dy}{d\theta}=e^r \sin(\theta)+e^r \cos(\theta)$$ Then after squaring both equations I got $2e^{2r}$, but how do I find the partial with respect to $x$ and $y$ when both variables are expressed in another set of variables?

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By the chain rule

$$ \frac{\partial z}{\partial r} =\frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r} $$

and

$$ \frac{\partial z}{\partial \theta} =\frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta}. $$

Substitute the appropriate partial derivatives, use the binomial theorem to expand the squares and use Pythagoras' trig identity to simplify. This will yield the required result.

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