I'm trying to tackle the following question
Let $f:\Bbb{R}^2\to\Bbb{R} \ , \ (x_0,y_0)\in\Bbb{R}^2 \ , \ \underline{u}=(u_1,u_2)\in \Bbb{R}^2$ where $\underline{u}$ is unit vector.
Let $g(t)=f(x_0+u_1t,y_0+u_2t)$. Show that $D_{\underline{u}}f(x_0,y_0)=g'(0)$.
My try: First, I think that it should be given that $g$ is differntiable at $x=0$. Now, $$D_{\underline{u}}f(x_0,y_0)= \lim_{t\to 0}\frac{f\left((x_0,y_0)+t\underline{u}\right)-f(x_0,y_0)}{t}=\lim_{t\to 0}\frac {f(x_0+u_1t,y_0+u_2t)-f(x_0,y_0)}{t} \\ g'(0)=\lim_{t\to 0}\frac{g(0+t)-g(0)}{t}=\lim_{t\to 0}\frac{f\left(x_0+u_1t,y_0+u_2t\right)-f(x_0,y_0)}{t}$$hence, the claim is proven. Is my reasoning fine?
Please help, thank you!
For your first question, yes, your reasoning is fine.
For the second question, use the chain rule.
You have $g(t)=f({\bf r}(t))$ where ${\bf r}(t) = (x_0,y_0)+t{\bf u} = (x_0+tu_1,y_0+tu_2)$.
You are given that $f$ is differentiable. ${\bf r}$ is linear and thus differentiable. Thus $g$ is differentiable as well.
Notice that ${\bf r}'(t) = {\bf u} = (u_1,u_2)$. The chain rule states that $g'(t) = f_x({\bf r}(t)) u_1 + f_y({\bf r}(t)) u_2$.
Plugging in $t=0$ yields: $g'(0)=f_x({\bf r}(0))u_1+f_y({\bf r}(0))u_2 = f_x(x_0,y_0)u_1+f_y(x_0,y_0)u_2=\nabla f(x_0,y_0) \cdot {\bf u}$
You already know that $g'(0)=D_{\bf u}f(x_0,y_0)$ so done. :)