# total derivative and directional derivative

Let $$E$$ be a subset of $$\mathbb{R}^n$$, $$f : E \to \mathbb{R}^m$$ be a function, $$x_0$$ be an interior point of $$E$$, and let $$v$$ be a vector in $$\mathbb{R}^n$$. If $$f$$ is differentiable at $$x_0$$, then $$f$$ is also differentiable in the direction $$v$$ at $$x_0$$ and $$D_vf(x_0) = f'(x_0) v.$$

Total derivative: let $$L : \mathbb{R}^n \to \mathbb{R}^m$$ be a linear transformation. We say that $$f$$ is differentiable at $$x_0$$ with derivative $$L$$ if we have $$\lim_{x \to x_0; x \in E\setminus \{x_0\}} \frac{||f(x) -(f(x_0) +L(x-x_0))||}{||x-x_0||} = 0.$$ Here $$||x||$$ is the length of $$x$$: $$||(x_1, ..., x_n)|| = (x_1^2 + ... + x_n^2)^{1/2}$$.

Directional derivative: if the limit $$\lim_{t\to 0; t>0, x_0+tv \in E} \frac{f(x_0+tv) - f(x_0)}{t}$$ exists, we say that $$f$$ is differentiable in the direction $$v$$ at $$x_0$$, and we denote the above limit by $$D_vf(x_0)$$.

We can first see that since the total derivative exists when $$x \to x_0$$, it still exists when $$x_0 +tv \to x_0$$. Thus, by substituting this variable, $$\lim_{t \to 0; t>0, x_0 +tv \in E} \frac{||f(x_0 + tv) -(f(x_0) +L(tv))||}{||tv||} = 0.$$

I am stuck here, and I am trying to transform this into the definition of directional derivative. But, I don't know how to justify eliminating $$|| \cdot ||$$ and how to change the denominator $$||tv||$$ into $$t$$. How can I proceed from here?

It is simple. In the following I'm assuming $$v\ne0$$ fixed, and $$t\searrow0$$. Total differentiability of $$f$$ at $$x_0$$ then implies $${f(x_0+tv)-f(x_0)-f'(x_0)(tv)\over|tv|}\to0\qquad(t\searrow0)$$ (I have written $$f'(x_0)$$ instead of $$L$$). As $$|tv|=t|v|$$ and $$|v|\ne0$$ this is equivalent with $${f(x_0+tv)-f(x_0)-f'(x_0)(tv)\over t}\to0\qquad(t\searrow0)\ .$$ This implies $${f(x_0+tv)-f(x_0)\over t}\ \to\ \lim_{t\searrow0}{f'(x_0)(tv)\over t}=f'(x_0)\,v\ .$$
• Thanks for you answer. In my book, total differentiability of $f$ at $x_0$ is $\lim_{t \to 0; t>0, x_0 +tv \in E} \frac{||f(x_0 + tv) -(f(x_0) +L(tv))||}{||tv||} = 0.$ Is it equivalent to saying ${f(x_0+tv)-f(x_0)-f'(x_0)(tv)\over|tv|}\to0\qquad(t\searrow0)\$? Commented Apr 1, 2020 at 22:33
• I am just not sure how to remove $|| \cdot ||$. Commented Apr 1, 2020 at 22:35
• My $|$ is your $\|$. – Total differentiability means there is such an $L$. This $L$ is uniquely determined by $f$ and $x_0$. It can therefore be called $f'(x_0)$, or similar. Commented Apr 2, 2020 at 8:16