# Is directional derivative at any point coincides with the total derivative?

I read directional derivative few days ago.I also read total derivative of a function.Now I have totally confused with this two topics.

Suppose there is a function $f : \mathbb R^2 \longrightarrow \mathbb R$.Suppose directional derivative of $f$ at $(a,b)$ in the direction of $(u,v)$ exists.Then

$$\lim_{h \rightarrow 0} \frac {f(a + hu,b + hv) - f(a,b)} {h}$$ exists.

Now let us associate two functions $x : \mathbb R \longrightarrow \mathbb R$ and $y : \mathbb R \longrightarrow \mathbb R$ defined by $x (t) = a + (t - c)u$ and $y (t) = b + (t - c)v$ respectively where $c \in \mathbb R$.

Then we have by above limit

$$\lim_{h \rightarrow 0} \frac {f(x(c + h),y(c + h)) - f(x(c),y(c))} {h}$$

which is actually $\frac {df} {dt} \vert_(x(c),y(c))$ i.e. $\frac {df} {dt} \vert_(a,b)$.

I think there is some big error in my calculation.Otherwise directional derivative of $f$ at $(a,b)$ along any direction coincides with the total derivative.Which is not true.Because total derivative implies continuity where directional derivative doesn't.Please tell me what is my fault.This will help me a lot in understanding this concept.

• By 'total derivative' do you mean the usual (Fréchet) derivative? – copper.hat Dec 31 '16 at 5:28
• I mean @copper.hat $\frac {df (x,y)} {dt} = \frac {\partial f} {\partial x} . \frac {dx} {dt} + \frac {\partial f} {\partial y} . \frac {dy} {dt}$ to be the total derivative of $f$ w.r.t. $t$. – Arnab Chatterjee. Dec 31 '16 at 5:36
• But @copper.hat does the existence of the limit $$\lim_{h \rightarrow 0} \frac {f(x(c + h),y(c + h)) - f(x(c),y(c))} {h}$$ in my post not ensure that $f$ is continuous at $(x(c),y(c))$ i.e. at $(a,b)$.If the answer is 'no' then please explain. – Arnab Chatterjee. Dec 31 '16 at 7:38
• Whatever you call the derivative above, it does not imply continuity, except along the direction of interest (where it it differentiable along that line). However, one can construct a simply example on the plane of a function $f$ that has a directional derivative everywhere, but is not continuous at the origin. – copper.hat Dec 31 '16 at 7:48
• But @copper.hat if the above limit exists say this $l$ then for we can write it as $f(x(c + h),y (c + h)) - f(x(c),y(c)) = hl + {\epsilon} h$, where $\epsilon \rightarrow 0$ as $h \rightarrow 0$.It implies that $$\lim_{h \rightarrow 0} [f(x(h + c),y(h + c)) - f(x(c),y(c))] = 0$$.Does it not imply that $f$ is continuous at $(x(c),y(c))$?Please don't mind if my question is too silly.But I have to understand it clearly before going to further study. – Arnab Chatterjee. Dec 31 '16 at 8:27

There is a danger of slipping into a notational morass which I will try to avoid.

There is no error as far as I can see. I think you have confused yourself with notation. Hopefully this may help straighten things out.

The directional derivative of $f$ in the direction $d_0$ at the point $x_0$ is given by $df(x_0,d_0) = \lim_{t \to 0 } {f(x_0+td_0)-f(x_0) \over t}$. In my notation $x_0,d_0$ are in $\mathbb{R}^2$ (or $\mathbb{R}^n$, more generally).

Note that if we let $\phi(t) = f(x_0+td_0)$, then $\phi'(0) = df(x_0,d_0)$.

Returning to your notation, we have $x(s) = (a,b)+(s-c) (u,v)$, and we consider $\eta(t) = f(x(t+c))$ and note that $\eta(t) = f((a,b))+ t (u,v))$, hence $\eta'(0) = df((a,b),(u,v))$.