I have the problem that $f(x,y)=g(r,\theta)$, where $x=r\cosh(\theta)$ and $y=r\sinh(\theta)$. I am trying to find $\frac{\delta f}{\delta x}$ from this by using partial derivates.
So far I have an the expression:
$$\frac{\delta g}{\delta r}=\frac{\delta f}{\delta x}(\cosh(\theta)+\frac{\delta f}{\delta y}(\sinh(\theta))$$
and another one similar for $\frac{\delta g}{\delta r}$,
Also using $x^2-y^2=r^2$ and $\theta=\operatorname{tanh}^{-1}(\frac{y}{x})$, I also get get expressions for $\frac{\delta r}{\delta x},\frac{\delta r}{\delta y}, \frac{\delta \theta}{\delta x},\frac{\delta \theta}{\delta y}$.
Finally I also have an expression for:
$$\frac{\delta g}{\delta x}=\frac{\delta g}{\delta r}\cosh(\theta)+\frac{\delta g}{\delta \theta}\frac{-\sinh(\theta)}{r}$$
and again a similar one for $\frac{\delta g}{\delta y}$.
But now I am not sure how to go about going any further... to get $\frac{\delta f}{\delta x}$ could anyone shed any light! Thanks!
That is correct. Since you already know the different partial derivatives involving the coordinates just observe that $\frac{\partial}{\partial x}=\frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta}+\frac{\partial r}{\partial x}\frac{\partial}{\partial r}$ and apply it to $g$ to obtain the partial derivative of $f$ in terms of the other coordinates.