Compute deformation gradient in an elasticity problem In the problem "finite bending of an incompressible elastic block, discussed here at pg. 181 deformation map is, given $r = f(x)$ and $\theta = g(y)$ two functions which will be determined later:
$$\mathbf{x} = [f(x) \cos(g(y)), f(x) \sin(g(y))] = f(x) \bigl( \cos(g(y)) e_1 + \sin(g(y))e_2 \bigr)$$
The author says that the deformation gradient is $$F = f' e_r \otimes e_1 + (fg') e_\theta \otimes e_2$$
where $e_r = \cos(\theta) e_1 + \sin(\theta) e_2$ and $e_\theta=-\sin(\theta)e_1 + \cos(\theta)e_2$.
Question
I don't know the logical steps to write it in that form, i.e. using the dyads $e_r \otimes e_1$. I'll try to write here what I think is the rationale:

*

*First compute the 4 derivatives, everything w.r.t. the variables $x,y$. They are $$f' \cos(\theta), f'\sin(\theta), -fg' \sin, fg' \cos $$


*Now I know that $F$, the deformation gradient, maps lagrangian elements into eulerian elements, so the basis must be $e_r \otimes e_1$ and $e_\theta \otimes e_2$ but I don't know why the first term is $f'$ and the other one $fg'$. My guess is the following: they worked out all the partials w.r.t $x,y$: $$f' \cos e_1 \otimes e_1 + fg' \cos e_2 \otimes e_2 + f' \sin e_2 \otimes e_1 - fg'\sin e_1 \otimes e_2 $$
and then he noted:
$$f'(\cos e_1 + \sin e_2) \otimes e_1 = f' e_r \otimes e_1$$ and similarly for the other term
 A: We have
$$\mathbf{x} = f(x_0)\cos(g(y_0))\hat{\mathbf{e}}_1
+ f(x_0)\sin(g(y_0))\hat{\mathbf{e}}_2\, .$$
We can write the deformation gradient as
$$\mathbf{F} = \frac{\partial \mathbf{x}}{\partial x_0}\otimes \hat{\mathbf{e}}_1 + \frac{\partial \mathbf{x}}{\partial y_0}\otimes \hat{\mathbf{e}}_2\, ,$$
or
\begin{align}
\mathbf{F} &= [f' \cos(g)\hat{\mathbf{e}}_1 +
f' \sin(g)\hat{\mathbf{e}}_2]\otimes \hat{\mathbf{e}}_1
+ [- fg' \sin(g)\hat{\mathbf{e}}_1 +
fg' \cos(g)\hat{\mathbf{e}}_2]\otimes \hat{\mathbf{e}}_2\\
&= f'[\cos(g)\hat{\mathbf{e}}_1 +
\sin(g)\hat{\mathbf{e}}_2]\otimes \hat{\mathbf{e}}_1
+ fg'[-\sin(g)\hat{\mathbf{e}}_1 + \cos(g)\hat{\mathbf{e}}_2]\otimes \hat{\mathbf{e}}_2\, .
\end{align}
And we can identify in the brackets $\hat{\mathbf{e}}_r$ and $\hat{\mathbf{e}}_\theta$. Thus
$$\mathbf{F} = \frac{d f(x_0)}{d x_0}\hat{\mathbf{e}}_r\otimes \hat{\mathbf{e}}_1 + f(x_0) \frac{d g(y_0)}{d y_0}\hat{\mathbf{e}}_\theta\otimes \hat{\mathbf{e}}_2\, .$$
A: In the deformed configuration, the differential position vector between to material points is $$\mathbf{ds}=e_rdr+ e_{\theta}rd\theta=f'e_rdx+(fg')e_{\theta}dy$$ where x and y are the material coordinates of the same pair of material points in the undeformed configuration.  But, this is the same as $$\mathbf{ds}=(f'e_r\otimes e_1+(fg')e_{\theta}\otimes e_2)\centerdot (e_1dx+e_2dy)=\mathbf{F}\centerdot \mathbf{ds_0}$$
Incidentally, isn't this the way we express the transpose of the deformation gradient tensor, rather than the deformation gradient itself?
