# Definition of directional derivative

I am thinking about the directional derivative. I think that the easiest way how to express it is

$$\frac{\partial f(\mathbf{x})}{\partial \mathbf{v}} = \lim_{||v|| \rightarrow 0} \frac{f(\mathbf{x} + \mathbf{v})-f(\mathbf{x})}{||\mathbf{v}||},$$

but the directional derivative is usually defined as

$$\frac{\partial f(\mathbf{x})}{\partial \mathbf{v}} = \lim_{h \rightarrow 0} \frac{f(\mathbf{x} + h \mathbf{v})-f(\mathbf{x})}{h}.$$

Can you rigorously explain the transition from first and second definition?

Thanks!

Edit:

Just to make it clear.

The first definition is primarily wrong because the orientation of the directional vector is not fixed. So if I correct it like this (switching to conventional notation)

$$\nabla_{\mathbf{v}} f(\mathbf{x}) = \lim_{||\mathbf{v}|| \to 0} \frac{f(\mathbf{x} + ||\mathbf{v}||\mathbf{\hat{v}})-f(\mathbf{x})}{||\mathbf{v}||},$$

it makes a little bit more sense ($\mathbf{\hat{v}}$ denotes unit vector). BUT the norm allows to get close to zero just from right side ($||\mathbf{v}|| \to 0+$) and the limit makes sense even for vector reversal (~ negative norm). So we can actually use any scalar $h$ scaling the vector and write the derivative as

$$\nabla_{\mathbf{v}} f(\mathbf{x}) = \lim_{h \to 0} \frac{f(\mathbf{x} + h\mathbf{\hat{v}})-f(\mathbf{x})}{h}$$

or you can find it equivalently written as

$$\nabla_{\mathbf{v}} f(\mathbf{x}) = \lim_{h \to 0} \frac{f(\mathbf{x} + h\mathbf{v})-f(\mathbf{x})}{h||\mathbf{v}||}.$$

• If it is a "directional" derivative, in what direction is the derivative. If you wanted the derivative in the "direction" of $(1,1,2)$, how would your definition do that? Your definition doesn't restrict $v$ to be "in that direction." – Thomas Andrews Feb 28 '13 at 16:13
• As an aside, if you want to allow non-unit $\mathbf{v}$, you get a better operator if you don't normalize: for example, you want to have $$\nabla_{2 \mathbf{v}} f = 2 \nabla_{\mathbf{v}} f$$ An example of the benefit is that this version still satisfies the formula $$\nabla_{\mathbf{v}} f = \mathbf{v} \cdot \nabla f$$ – user14972 Mar 1 '13 at 9:03

Your definition is not "directional," since $v$ in your definition can be in any direction. Note, if you were just in $1$ dimension, the limit $$\lim_{\delta\to 0} \frac{f(x+\delta)-f(x)}{|\delta|}$$ isn't even defined when $f$ is normally differentiable, because when $\delta\to 0-$ the limit is $-f'(x)$and when $\delta\to 0+$ the limit is $f'(x)$.

Essentially, the "directional derivative" can be seen as taking $\mathbf x$ and $\mathbf v$ and defining a new function on the real numbers: $g(h) = f(\mathbf x + h\mathbf v)$. This $g$ computes the value of $f$ on the line through $\mathbf x$ in the direction of $\mathbf v$, and the directional derivative of $f$ at $\mathbf x$ in the direction $\mathbf v$ is defined as $g'(0)$.

• It's worse than that; his definition doesn't even make sense from a notational viewpoint! – Noldorin Feb 28 '13 at 16:14
• thanks, now I understand ... :-) – OukiDouki Feb 28 '13 at 16:30
• Please, look at the edit above. I want to know if I get it right. (Sorry for asking to obvious/stupid question.) – OukiDouki Mar 1 '13 at 9:01
• That still doesn't deal with the case of $h<0$. You are allowing, which is not the same as my definition. For example, in your definition, $f(\mathbf v)=|\mathbf v|$ has direction derivative at $\mathbf v=0$, but in my definition, it does not. That's because you have $|v|\to 0$. Also, your definition is confusing since nothing make $\mathbf v$ be in the same vector as $\hat{\mathbf{v}}$ - you are essentially using $\hat{\mathbf{v}}$ in your limit as a stand-in for a positive real - you might as well say $$\lim_{h\to 0+}$$ – Thomas Andrews Mar 1 '13 at 12:14

Your first definition makes no sense really, since you're using $\Delta \mathbf{x}$ simultaneously as a parameter and variable (which you're taking a limit over). The second (correct) definition thus uses a separate variable over which the limit is taken, used to scale the positional vector.

• sorry, corrected – OukiDouki Feb 28 '13 at 15:57
• It still doesn't make sense I'm afraid; the RHS is what doesn't work, as explained above. As for the LHS, you're using unconventional notation (the nabla symbol is almost always used), but I get what you mean. – Noldorin Feb 28 '13 at 16:02