Let $f:\Bbb{R}^2\to \Bbb{R}$ such that the partial derivatives of $f$ exist. Let $x(t),y(t):\Bbb{R}\to\Bbb{R}$ be differentiable functions and let $g(t)=f\left(x(t),y(t)\right)$. Show that $g$ is differentiable and $g'(t)=\nabla f\left(x(t),y(t)\right)\cdot{\left(x'(t),y'(t)\right)}$.
My try: We want to show that $g$ differentiable, hence we need to show that the limit $\displaystyle \lim_{h\to 0}\frac{g(t+h)-g(t)}{h}$ exists. Using the definition of $g$ we need to show that $\displaystyle \lim_{h\to 0}\frac{f\left(x(t+h),y(t+h)\right)-f\left(x(t),y(t)\right)}{h}$ exists. How should I continue?
Use the chain rule. Referring to your previous question...
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. :)