On the definition of the directional derivative In the multivariable calculus course I took the directional derivative of a multivariable function $f(x,y)$ at $(a,b)$ in the direction of the vector $\vec{s}$ was defined as the following:
$$f_s(a,b) = \vec{\nabla f} \cdot \vec{u_s}$$
where $\vec{u_s}$ is the unit vector in the same direction of $\vec{s}$. Now I have come across the following definition:
$$\frac{d}{d\alpha} f(\vec{v} + \alpha\vec{s})$$
evaluated at $\alpha = 0$ $(\vec{v}$ is supposed to be the vector at which the derivative is evaluated). I am struggling to see why the two definition are equal.
 A: The definitions are not equivalent. As far as I know, the usual definition is the second one:

Definition: The directional derivative of $f$ at $p$ in the direction of the unit vector $\vec{s}$ is the derivative at $0$ of the function $\varphi(t)=f(p+t\vec{s})$

The first "definition" in the question is actually a property that is not equivalent to the definition. It is true for example if $f$ is differentiable at $p$:

Theorem: If $f$ is differentiable at $p$, then $f$ has directional derivatives at $p$ in every direction $\vec{s}$ and $f_{\vec{s}}(p)=\nabla f(p)\cdot \vec{s}$.

The proof is simply a consequence of the Chain Rule for differentiable functions. If $f$ has all partial derivatives (hence has a gradient) but is not differentiable, things get messy. The RHS is defined but it is possible that the LHS does not exist for some direction $\vec{s}$. Worse, it is possible that $f_{\vec{s}}(p)$ exists but is not equal to $\nabla f(p)\cdot \vec{s}$!
Some interesting examples (at $p=(0,0)$):

*

*$f(x,y)=0$ if $x=0$ or $y=0$, and $f(x,y)=1$ otherwise.

*$f(x,y)=\frac{y^2}{x}$ if $x\neq 0$ and $f(0,y)=0$.

Those examples are classic Calculus/analysis textbooks material. In some contexts (differential geometry?), we don't care about these kinds of pathological examples and consider only differentiable functions, so the formula is always true.
A: By the chain rule you have
\begin{align*}
\frac{d}{d\alpha}f(v+\alpha s)_{|\alpha=0}
&=\nabla f(v+\alpha s_{|\alpha=0})\cdot \frac{d}{d\alpha}(v+\alpha s)_{|\alpha=0}\\
&=\nabla f(v)\cdot s
\end{align*}
which should solve your problem.
On the other hand observe that $\frac{d}{d\alpha}f(v+\alpha s)_{|\alpha=0}$ is the directional derivative of $f$ in the direction of the vector $s$ evalued at point $v$, that is $\partial_sf(v)$.
This leads to the well known formula in multivariable calculus:
$$
\nabla f(v)\cdot s=\partial_sf(v)\;.
$$
Here $f:U\to\Bbb R$ where $U$ is open in $\Bbb R^n$; note the switching role the elements of $\Bbb R^n$ have: both vectors and points, according on what needed (namely: both $v$ and $s$ are element of $\Bbb R^n$, but $v$ is seen as a point, while $s$ as vector, since we are moving in its direction to reach the derivative).
