Given $f:\mathbb{R}^2 \to \mathbb{R}^2$ is a map $f(x,y)=(u(x,y),v(x,y))$ and $\alpha=(\alpha_1,\alpha_2)$ is a point, then how does one show that $f$ is differentiable (or not) in the direction $v=(v_1,v_2)$?
For example, what if $f(x,y)=(e^x \cos y,e^x \sin y)$ and $\alpha=(0, \pi/3)$ and $v=(9,12)$? Given that this function apparently has a directional derivative at $\alpha$, how does one go about to show this using the definition?
Is it to show that that the Jacobian is defined at $\alpha$ and then multiply this by the direction vector $v$ (which has been normalized)?
The definition of differentiation is as follows \begin{equation} \lim_{h \to 0}\frac{|f(\overline{x}+\overline{h})-f(\overline{x})-A\overline{h}|}{|\overline{h}|}=0 \end{equation}
I have calculated \begin{equation} A=\begin{pmatrix} e^x \cos y &-e^x\sin y \\ e^x \sin y & e^x \cos y \end{pmatrix} \end{equation} and so at $\alpha =(0, \pi/3)$ \begin{equation} A=\begin{pmatrix} 1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{pmatrix} \end{equation} Therefore, if \begin{align} f(\overline{\alpha}+\overline{h})&=(e^{0+h}\cos (\pi/3+h), e^{0+h}\sin(\pi /3 +h)) \\ &=(e^h [\cos (\pi/3)\cos(h)-\sin(\pi/3)\sin(h)], e^h[\cos(\pi/3)\sin (h)+\cos (h)\sin(\pi/3]) \\ & =\end{align}