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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.

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  • $\begingroup$ Try using the chain rule to evaluate $\frac{d}{d\alpha}f(\vec v + \alpha \vec s)$ at $\alpha = 0$. $\endgroup$ – littleO Oct 6 '20 at 0:57
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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 $p=(0,0)$):

  1. $f(x,y)=0$ if $x=0$ or $y=0$, and $f(x,y)=1$ otherwise.
  2. $f(x,y)=\frac{y^2}{x}$ 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.

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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).

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  • $\begingroup$ But the Chain Rule is not true if the functions are not differentiable. $\endgroup$ – Taladris Oct 6 '20 at 1:01
  • $\begingroup$ I am pretty sure that a person in trouble with multivariable class doesn't concieve anything which is not differentiable. Otherwise it would have been stated explicitly. $\endgroup$ – Joe Oct 6 '20 at 1:06
  • $\begingroup$ And you solve the OP's struggle by giving him information that are factually wrong? That costs nothing to be precise and saying that it works when $f$ is differentiable, but things get more complicated in other contexts. $\endgroup$ – Taladris Oct 6 '20 at 1:15

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