Directional derivative understanding [Beginning multivariable question.] I have just been introduced to a theorem that says $$D_uf(x)=\nabla f(x)\cdot u.$$ So in the two-dimensional case, $$\nabla f(x,y)= \langle f_x(x,y),f_y(x,y)\rangle \cdot \langle a,b\rangle$$
I don't really understand this. It seems to me that $f_x$ could be 0 and $f_y$ could be 0, there could still be a nonzero derivative in the direction halfway between the $x$ and $y$ axes. 
My intuition is that such a situation would violate the theorem -- so I must be misunderstanding something. Where am I going wrong?
 A: First, in cartesian coordinates we have a quantity known as gradient defined the following way:
$$ \nabla F = < \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}>$$
Now, it may be noted that:
$$ F(x+ \Delta x, y+ \Delta y) = F(x, y + \Delta y) +  \Delta x \frac{\partial F}{\partial x}(x,y+\Delta y)+\text{higher order terms in x } \tag{1}$$
And, further we can write:
$$ F(x,y + \Delta y) = F(x,y) + \frac{\partial F}{\partial y}(x,y)\Delta y + +\text{higher order terms in y}  \tag{2}$$
This leads to:
$$ F(x+ \Delta x, y+ \Delta y) = F(x,y) + \frac{\partial F}{\partial y}(x,y)\Delta y + +\text{higher order terms in y} +  \Delta x \frac{\partial F}{\partial x}(x,y+\Delta y)+\text{higher order terms} \tag{1}$$
Or,
$$ F(x+ \Delta x, y+ \Delta y) \approx F(x,y) + \frac{\partial F}{\partial y}(x,y)\Delta y +\Delta x \frac{\partial F}{\partial x}(x,y+\Delta y)\tag{3}$$
It may be noted that for small enough $\Delta x$ and $\Delta y$ we can write: $$ \Delta x \frac{\partial F}{\partial x} (x,y+\Delta y) = \frac{\partial F}{\partial y}(x,y) \Delta y $$
This leads to:
$$ F(x+ \Delta x, y+ \Delta y) \approx F(x,y) + \frac{\partial F}{\partial y}(x,y)\Delta y +\Delta x \frac{\partial F}{\partial x}(x,y)\tag{4}$$
Or,
$$ F(x+ \Delta x, y+ \Delta y) \approx F(x,y) + \nabla F \cdot <\Delta x, \Delta y>\tag{4}$$
(1) and (2) were based on Taylor series/ linear approximation to F. I have written about it here
A: Firstly the statement $\nabla f(x,y)=(f_x(x,y),f_y(x,y))\cdot(a,b)$ is incorrect. $\nabla f(x,y)$ is called the gradient of $f$. It exists whenever $f$ is a function $\mathbb{R^n}\to\mathbb{R}$ and all of $f$'s partial derivatives exist. For multivariate functions it is always a vector, for single variate functions it reduces to $f'$. For $f:\mathbb{R}^2\to\mathbb{R}$ it is defined $\nabla f(x,y)=(f_x(x,y),f_y(x,y))$. If you evaluate $\nabla f$ at a point $(x,y)$ you will either get a non-zero vector pointing in the direction of maximum increase of $f$ at $(x,y)$, or a stationary point where $\nabla f=\mathbf{0}$ (if you haven't already you will learn how the concept of a stationary point generalises to the multivariable case.). 
As an example, consider $f(x,y)=x^2y^2$ (plot it in WolframAlpha.) The gradient of $f(x,y)=x^2y^2$ is $\nabla f(x,y)=(2xy^2,2x^2y)$. Stationary points exist whenever $x$ or $y$ are zero. We have $\nabla f(1,1) = (2,2)$, which points in the direction of maximum increase of $f$ at $(1,1)$, or up the "steepest slope" of $f$ at $(1,1)$. 
The directional derivative of a function $f:\mathbb{R^n}\to\mathbb{R}$ in the direction $\mathbf{a}\in\mathbb{R}^n$ at the point $\mathbf{x}\in\mathbb{R}^n$ is defined as 
$$D_\mathbf{a} f(\mathbf{x}) = \lim_{h\to 0} \frac{f(\mathbf{x}+h\mathbf{a})-f(\mathbf{x})}{h}.$$
It is clearly a different object to the gradient.
As an example, consider the same function $f$ from before. The directional derivative is,
\begin{align}
D_{(a,b)} f(x,y)&= \lim_{h\to 0} \frac{ (x+ha)^2(y+hb)^2-x^2y^2}{h} \\
&= \lim_{h\to 0}\, (ay + bx + abh)(2xy + ahy + bhx + abh^2) \\
&= 2xy(ay +bx)
\end{align}
Evaluating at $(1,1)$ in the direction $(-1,-1)$ gives $-4$.
The theorem you referred to in your question requires that $f$ be differentiable. The terminology is confusing because this has a technical meaning you will learn later. For now, just remember that the theorem only applies if all of $f$'s partial derivatives exist and are continuous. If this is the case, then the theorem applies. Thus it is impossible to find a function with continuous partial derivatives equal to zero along each axis at a given point, but with a non zero directional derivative in some other direction at that point. 
With our example function $f$, we had $\frac{\partial f}{\partial x}=2xy^2$ and $\frac{\partial f}{\partial y}=2x^2y$, both of which are continuous. We had $\nabla f(1,1)= (2,2)$ and $\nabla f(1,1)\cdot (-1,-1)=-4$ which was $D_{(-1,-1)} f(1,1)$ from before!
