Q:

Help me prove: directional derivative in terms of partial derivative

Given a function $f(x, y)$ which has partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$, and whose partial derivatives are also differentiable, and    $\vec r = (x_r, y_r)$, I am trying to prove that the derivative of $f$ along $\vec r$ is:
$Df_r (x,y) = \lim_{h\to 0} \dfrac{f(x+hx_r, y+hy_r)-f(x,y)}{h} = x_r \dfrac{\partial f}{\partial x} + y_r \dfrac{\partial f}{\partial y} = D$
I am trying to prove this by showing that for any given $\epsilon$, I can find a $\delta$ such that for all $h < \delta$, the actual rate-of-change is within $\epsilon$ of $D$.
We can write $f(x+hx_r, y+hy_r)-f(x,y) = A + B$, where $A = f(x+hx_r, y)-f(x,y)$ and $B = f(x+hx_r, y+hy_r)-f(x+hx_r,y)$. I wish to make $|A + B| \le |A| + |B| \le (D + \epsilon)h$ ...[1].  
Since $\frac{\partial f}{\partial x}$ exists, I can pick an $h$ to bring |A| within a factor of any $\epsilon_1$ to $x_r\frac{\partial f}{\partial x} h$. I can make the $h$ smaller, if necesssary, to bring $\frac{\partial f}{\partial y}(x+hx_r, y)$ within $\epsilon_2$ of $\frac{\partial f}{\partial y}(x, y)$. I then wish to bring $|B|$ sufficiently close to $y_r \frac{\partial f}{\partial y} h$, so that [1] is achieved. 
I am getting stuck at the last step. Because $\frac{\partial f}{\partial y}$ exists everywhere, for all $\epsilon_3$ there exists an $\delta_3$ such that for all $h_3 < \delta_3$, $|f(x+hx_r, y+h_3 y_r) - f(x+hx_r, y)| \le (y_r \frac{\partial f}{\partial y} h_r + \epsilon_3)h_3$. In my situation, however, $h_3$ is not a free variable, but fixed: $h_3 = h$. No matter how small I make my $h$, $\delta_3$ might be too small.

A:

In order to have this equality $Df_r (x,y) =\displaystyle \lim_{h\to 0} \dfrac{f(x+hx_r, y+hy_r)-f(x,y)}{h} = x_r \dfrac{\partial f}{\partial x} + y_r \dfrac{\partial f}{\partial y} =\nabla f.\vec r= D\qquad (1)$, the function $f$ has to be differentiable at $(x,y)$, which has already been given. Recall that if $f$ is differentiable at $(x, y)$, there are functions $\varepsilon_1=\varepsilon_1(\Delta x, \Delta y)$ and $\varepsilon_2=\varepsilon_2(\Delta x, \Delta y)$ with $\varepsilon_1\rightarrow 0$ and $\varepsilon_2\rightarrow 0$ such that $$f(x+\Delta x, y+\Delta y)-f(x, y)=f_x(x, y)\Delta x+f_y(x, y)\Delta y+\varepsilon_1\Delta x+\varepsilon_2\Delta y.\quad (2)$$
Note that $\vec r$ is a unit vector, that is, $x_r^2+y_r^2=1$. With the $\varepsilon$-$\delta$ language, (2) can be expressed as: for any given $\varepsilon>0$ can find 
 a $\delta>0$ such that $\sqrt{(\Delta x)^2+(\Delta y)^2}<\delta\Longrightarrow$ 
$$\bigg|\frac{f(x+\Delta x, y+\Delta y)-f(x, y)-f_x(x, y)\Delta x-f_y(x, y)\Delta y}{\sqrt{(\Delta x)^2+(\Delta y)^2}}\bigg|<\varepsilon.$$
Now setting $\Delta x=x_rh$, and $\Delta y=y_rh$ we have
$|h|=\sqrt{(x_r^2+y_r^2)}|h|=\sqrt{(x_r^2+y_r^2)h^2}=\sqrt{(\Delta x)^2+(\Delta y)^2}<\delta\Longrightarrow$ 
 $$\bigg|\frac{f(x+x_rh, y+y_rh)-f(x, y)-f_x(x, y)x_rh-f_y(x, y)y_rh}{h}\bigg|<\varepsilon.$$
or $$\bigg|\frac{f(x+x_rh, y+y_rh)-f(x, y)}{h}-f_x(x, y)x_r-f_y(x, y)y_r\bigg|<\varepsilon.$$
This implies equality (1).

