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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?

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1 Answer 1

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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

$$\lim_{t\to0}\frac{f(a+tv)-f(a)-\mathrm Df(a)(tv)}{t}=0.$$

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