Proof of directional derivative explanation

Question

From Stewart's Calculus the part of the proof states that:
If we define a function g of the single variable h by $$g(h) = f(x_0+ha,y_0+ha)$$ then by the definition of the derivative $$\lim_{{h \to 0}}\frac{g(x+h)-g(x)}{h}$$ we have $$g'(0) = \lim_{{ h \to 0}}\frac{g(h)-g(0)}{h}$$ but then author replaces $$g(h)$$ by $$f(x_0+ha,y_0+ha)$$ so we receive $$\lim_{{ h \to 0}}\frac{f(x_0+ha,y_0+ha)-f(x_0,y_0)}{h}$$ So the question is why 'h' in derivative of g(h) both in numerator and denominator equals to 'h' in $$f(x_0+ha,y_0+ha)$$ When 'h' in g(h) represents projection vector length, but 'h' in derivative of g(h) represents small changes in input?

Answer 1

$h$ is the free variable of $g(\cdot)$ and of $f\left(x_0+a\times(\cdot), y_0+a\times(\cdot)\right)$.

Let's for argument's sake define these functions independently of each other and we'll give them different free-variable names, viz. $g(h_1)$, $f\left(x+a\times(h_2), y+a\times(h_2)\right)$, $e(x,h_3)\triangleq g(x+h_3)$. Then by the constraint $g(h)=f(x_0+ha,y_0+ha)\,\forall h$ we have $h_1=h_2$.

Now by our definition, $e(x,h_3)=g(x+h_3)=f(x_0+ax+ah_3, y_0+ax+ah_3)$ where $h_1=h_2=x+h_3$ and

$$\begin{align}g'(x)&=\underset{h_3\to 0}{\lim}\frac{e(x,h_3)-g(x)}{h_3}\\ g'(0)&=\underset{h_3\to 0}{\lim}\frac{e(0,h_3)-g(0)}{h_3} \quad \text{$x+h_3=0\implies x=-h_3,\quad $so}\\ g'(0)&=\underset{h_3\to 0}{\lim}\frac{e(0,h_3)-e(0,0)}{h_3}\\ &=\underset{h_3\to 0}{\lim}\frac{f(x_0+a\cdot 0+ah_3, y_0+a\cdot 0+ah_3)-f(x_0+a\cdot 0+a\cdot 0, y_0+a\cdot 0+a\cdot 0)}{h_3}\\ &=\underset{h_3\to 0}{\lim}\frac{f(x_0+ah_3, y_0+ah_3)-f(x_0, y_0)}{h_3} \end{align}$$

In your post, all $\Delta h$ should be replaced with $h$