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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

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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 ..

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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 $\mathbf{x}$.

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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 $c'(0)=u$.

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