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?

Thanks in advance.


1 Answer 1


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 $${f(x_0+tv)-f(x_0)\over t}\ \to\ \lim_{t\searrow0}{f'(x_0)(tv)\over t}=f'(x_0)\,v\ .$$

  • $\begingroup$ 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)\ $? $\endgroup$
    – shk910
    Commented Apr 1, 2020 at 22:33
  • $\begingroup$ I am just not sure how to remove $|| \cdot ||$. $\endgroup$
    – shk910
    Commented Apr 1, 2020 at 22:35
  • $\begingroup$ 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. $\endgroup$ Commented Apr 2, 2020 at 8:16

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