# The formal definition of directional derivatives

So, let's assume that I have a line $$y=mx$$ in cartesian systems and that I want to find the directional derivative of the function $$f(x,y)$$ along this line. The position vector $$\vec{r}$$ and the unit vector in this direction would be $$\vec{v}=\frac{1}{\sqrt{1+m^2}}\pmatrix{1\\m}\quad \quad \vec{r}=\pmatrix{x\\y}$$ Formally the directional derivative in this direction would be defined as $$\partial_vf=\lim_{\lambda\to 0}\frac{f(\vec{r}+\lambda\vec{v})-f(\vec{r})}{\lambda}=\lim_{\lambda\to0}\frac{f(x+\frac{\lambda}{\sqrt{1+m^2}},y+\frac{\lambda m}{\sqrt{1+m^2}})-f(x,y)}{\lambda}$$ Now, some authors define directional derivatives as $$\partial_vf=\nabla f\cdot\vec{v}$$, hence in accordance to them $$\partial_v f=\lim_{\lambda\to0}\left[\frac{f(x+\lambda,y)-f(x,y)}{\lambda}.\frac{1}{\sqrt{1+m^2}}-\frac{f(x,y+\lambda)-f(x,y)}{\lambda}.\frac{m}{\sqrt{1+m^2}}\right]$$ Where am I going wrong? Kindly suggest a direction...

• What makes you think that the two limits aren’t equal? – amd May 26 '19 at 22:06
• The second “definition” is a convenient result that holds for differentiable $f$. – amd May 26 '19 at 22:06
• @amd, but is the second definition an identity? In other words, should I take it seriously? And is it ALWAYS an identity if $f$ is differentiable? – ubuntu_noob May 26 '19 at 22:10

Recall the Taylor expansion of a differentiable function in $$\mathbb{R}^n$$
$$f(\mathbf{x}+\mathbf{a})=f(\mathbf{x})+\mathbf{a}\cdot\mathbb{\nabla} f(\mathbf{x})+\mathcal{O}(|\mathbf{a}|^2)$$
use $$\mathbf{a}=\lambda\mathbf{v}$$ to get
$$\frac{f(\mathbf{x-\lambda v})-f(\mathbf{x})}{\lambda}=\mathbf{v}\cdot\mathbb{\nabla} f(\mathbf{x}) + \mathcal{O}(\lambda^2)$$
take the limit $$\lambda\rightarrow 0$$ to get the result