Directional Derivatives - Multivariable Calculus Currently studying directional derivatives. I have the definition of directional derivatives in terms of limits, how I don't understand how we get from the limit stage to showing that $\color{blue}{{\rm grad}(f)\cdot {\bf u}}$ (which is the direction vector) is equal to the directional derivative. 
I appreciate any help, please explain it to me as easy as possible.
Thank You
 A: If you are looking for an intuitive  sketch, take a 2D surface $z=f(x,y)$.   Determine the tangent plane in $(x_0,y_0)$.
Take a vectorial increment $\Delta {\bf t} =(\Delta t_x, \Delta t_y)=\Delta|\bf t |(cos\alpha,sin\alpha)$.
Then it is clear that if you move $\Delta t_x$ in the $x$ direction and $\Delta t_y$ in the $y$ direction, $\Delta f(x,y)$ will be ..
A: Consider a multivariate function $f:\mathbb{R}^n \rightarrow \mathbb{R}$, and suppose we want the directional derivative in an arbitrary direction $\mathbf{u} \in \mathbb{R}^n$ (with $||\mathbf{u}|| = 1$) at a point $\mathbf{x} \in \mathbb{R}^n$.  To facilitate our analysis, let $\mathbf{u}_k = (u_1, \cdots, u_k, 0, \cdots, 0) \in \mathbb{R}^n$ be the directon vector with all but its first $k$ elements set to zero.  If $f$ is differentiable at $\mathbf{x}$ (which is required for the result) then we have:
$$\begin{equation} \begin{aligned}
\nabla_\mathbf{u} f(\mathbf{x}) 
&\equiv \lim_{h \rightarrow 0} \frac{1}{h} [ f(\mathbf{x} + h \mathbf{u}) - f(\mathbf{x}) ] \\[8pt]
&= \lim_{h \rightarrow 0} \frac{1}{h} \Big[ \sum_{k=1}^n (f(\mathbf{x} + h \mathbf{u}_k) - f(\mathbf{x} + h \mathbf{u}_{k-1})) \Big] \\[8pt]
&= \sum_{k=1}^n  \lim_{h \rightarrow 0} \frac{1}{h} \Big[ f(\mathbf{x} + h \mathbf{u}_k) - f(\mathbf{x} + h \mathbf{u}_{k-1}) \Big] \\[8pt]
&= \sum_{k=1}^n  \lim_{h \rightarrow 0} \frac{1}{h} \Big[ f(\mathbf{x} + h \mathbf{u}_{k-1} + h u_k \mathbf{e}_k) - f(\mathbf{x} + h \mathbf{u}_{k-1}) \Big] \\[8pt]
&= \sum_{k=1}^n u_k \lim_{h u_k \rightarrow 0} \frac{1}{h u_k} \Big[ f(\mathbf{x} + h \mathbf{u}_{k-1} + h u_k \mathbf{e}_k) - f(\mathbf{x} + h \mathbf{u}_{k-1}) \Big] \\[8pt]
&= \sum_{k=1}^n u_k \cdot \frac{\partial f}{\partial x_k}(\mathbf{x}) \\[8pt]
&= \mathbf{u} \cdot \nabla f(\mathbf{x}). \\[8pt]
\end{aligned} \end{equation}$$
This working is an application of the multivariate chain rule.  The step from the fifth to the sixth line makes use of the assumption that $f$ is differentiable at $\mathbf{x}$.
A: Taking the composition $c(t)=p+tu$ and $F(t)=f\circ c(t)$ assigns the measure $f$ on the curve $c(t)$, then the directional derivative is 
$F'(t)={\rm grad} f(c(t))\cdot c'(t)$. But $c'(t)=u$ then
$$\frac{dF}{dt}=\frac{\partial f}{\partial u}={\rm grad} f(c(t))\cdot u.$$
Note that $c(0)=p$ and $c'(0)=u$. 
