# The directional derivative is the Jacobian

Let $$f:\mathbb{R}^m \rightarrow \mathbb{R}^n$$ differentiable at $$x$$ and $$v$$ an unit vector in $$\mathbb{R}^m$$. I've always used the formula that the directional derivative in $$x$$ by $$v$$ is simply the Jacobian times the $$v$$. However, I could never figure it out how to prove it.

$$\frac{\partial f}{\partial v}(x) = Df(x) v$$

In this question, the answer simply assumes that

$$\lim_{t\to0}\frac{\bigl\lVert f(a+tv)-f(a)-Df(a)(tv)\bigr\rVert}{\lVert tv\rVert}=\lim_{t\to0}\frac{f(a+tv)-f(a)-tDf(a)(v)}t$$

But it apparently ignores the triangular inequality in the given norm.

What am I not seeing here? How can I prove it?

It just uses that $$\Vert tv\Vert=\vert t\vert\Vert v\Vert$$, so we have
$$\frac{\Vert f(a+tv)-f(a)-\mathrm Df(a)(tv)\Vert}{\Vert tv\Vert}=\frac{1}{\Vert v\Vert}\left\Vert\frac{f(a+tv)-f(a)-\mathrm Df(a)(tv)}{t}\right\Vert.$$
Now the condition is that the limit of this expression is $$0$$. This means that
$$\lim_{t\to0}\left\Vert\frac{f(a+tv)-f(a)-\mathrm Df(a)(tv)}{t}\right\Vert=0,$$
since $$\frac{1}{\Vert v\Vert}$$ is constant. And a function goes to zero if and only if its norm goes to zero, since $$f(x)\to a$$ means $$\Vert f(x)-a\Vert\to0$$ by definition. So this is equivalent to
$$\lim_{t\to0}\frac{f(a+tv)-f(a)-\mathrm Df(a)(tv)}{t}=0.$$