Directional derivative.

Question

Let us assume that we have the function $z=f(x,y)$. Now in one variable the derivative $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$, we know we cannot do $+2h$, since there is a difference. So now i am confused as to how we pick that constant in front of $h$ in two variables. Say we have $z=f(x,y)$ and there is a direction vector $u=<a,b>$ and it is a unit vector but $2u=<2a,2b>$ is also a vector that correctly points to the direction. So according to my textbook the directional derivative is $\lim{h\to 0}\frac{f(x_0+ha,y_0+hb)-f(x_0,y_0)}{h}$, however, how do we know this is the "right" constants? Why not $2ha,2hb$.

Answer 1

Note that since $h\to 0 \implies 2h \to 0$

$$f'(x)=\lim_{h\to 0}\frac{f(x+2h)-f(x)}{2h}$$

For two (several) variables often the directional derivative are defined assuming $\vec u$ as a unit vector (as for partial derivatives), otherwise more in general we have that:

$$\frac{\partial f}{\partial \lambda \vec u}=\lim_{h\to 0}\lambda\frac{f(x_0+\lambda ah,y_0+\lambda bh)-f(x_0,y_0)}{\lambda h}=\lambda \frac{\partial f}{\partial \vec u}$$