Trouble reading directional derivative proof I'm reading Vector Calculus from http://mecmath.net/. This is a free PDF book for students of Calculus III. In section 2.4 (page 78) it introduces the directional derivative and theorem 2.2:

So far so good and I get the intuition. I am having trouble following the proof of the theorem though. Please see following picture:

I understand the first equality; a partial w.r.t. $y$ is being taken so $x$ is being held constant at $a + hv_1$ therefore it can be considered the derivative of a function $g(y) = f(a+hv_1,y)$.
I'm confused where $\alpha$ goes after the 2nd equals sign. I'm confused that $g\prime$ is claimed to be equal to some difference quotient and not the limit of the difference quotient. If anyone could talk me through what's going on here I'd be grateful.
 A: This seems to be a very over complicated way of proving $D_{\underline{u}}f(x_{0},y_{0}) = af_{x}(x_{0},y_{0})+ bf_{y}(x_{0},y_{0})$ where $(a,b)$ is the unit vector associated with $\underline{u}$ i.e. $(a,b)$ =  $\frac{\underline{u}}{\left \| \underline{u}  \right \|}$. We will prove this in a way that is much easier to understand.
To do this we do define $g:\mathbb{R} \to \mathbb{R}$ by $g(h) = f(x_{0} +ah,y_{0} +bh)$.
So, from our difference quotient we get 
$g'(0)$ = $\lim_{h \to 0}\frac{g(h) - g(0)}{h} = \lim_{h \to 0}\frac{f(x_{0} + ah,y_{0} + bh) - f(x_{0},y_{0})} {h} = D_{u}f(x_{0},y_{0})$
But, using the chain rule we get $g'(h) = \frac{\mathrm{d} g}{\mathrm{d} h} = \frac{\partial f}{\partial x}\frac{\mathrm{d} x}{\mathrm{d} h} + \frac{\partial f}{\partial y}\frac{\mathrm{d} y}{\mathrm{d} h}$ and we note that $x = x_{0} + ah$, $y = y_{0} + bh$ so $\frac{\mathrm{d} x}{\mathrm{d} h} = a $ and $\frac{\mathrm{d} y}{\mathrm{d} h} = b$
Then we have $g'(h) = a \frac{\partial f}{\partial x} + b \frac{\partial f}{\partial y}$ and so we evaluate this at $h = 0 \implies (x,y) = (x_{0}, y_{0})$ and equate our different expressions to get $D_{\underline{u}}f(x_{0},y_{0}) = a \frac{\partial f}{\partial x} (x_{0}, y_{0}) + b \frac{\partial f}{\partial y}(x_{0}, y_{0})$ which is what we wanted.
