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As the topic, let $\mathbf{u}=u\hat{i}+v\hat{j}$.

Why $$D_ug(0,0) = \lim_{h\rightarrow0}\frac{g(hu,hv)-g(0,0)}{h}$$ while

$$Dg(0,0)= \lim_{(h,k)\rightarrow (0,0)}\frac{g(h,k)-L(0,0)}{\sqrt{h^2+k^2}},$$we don't need to use the linearization in the case of directional derivatives.

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Some notes on formatting: To type a vector in bold you can use \mathbf{ }. For the unit vectors you can use \hat{ }. Also use \lim for limits. –  user38268 Jan 15 '12 at 14:13
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1 Answer 1

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I will assume that $L(0,0) = g(0,0)$; otherwise, the error in the second definition will be elsewhere, but I will need the definition of $L$ to pinpoint it.

The problem is that the second definition is false (in that it does not match with the usual definition of the derivative). For instance, if we take $g(h,k) = h$, we get:

$$ Dg (0,0) = \lim_{(h,k) \to (0,0)} \frac{h}{\sqrt{h^2+k^2}},$$

but this limit does not exist: if $h=0$ and $k$ goes to $0$, then the limit is $0$, but if $k=0$ and $h$ goes to $0$ then this limit is $\pm 1$. So, with this definition, even this function $g$ would not be differentiable!

The definition of the derivative $Dg (0,0)$ of $g$ in $(0,0)$ is that it is (if it exists) a linear operator from $\mathbb{R}^2$ to $\mathbb{R}$, such that for all vectors $\mathbf{u}$:

$$ \lim_{(h,k) \to (0,0))} \frac{g(h,k) - g(0,0) - Dg (0,0) \mathbf{u}}{\sqrt{h^2+k^2}} = 0,$$

or equivalently:

$$g(h,k) =_{(0,0)} g(0,0) + Dg (0,0) (h,k) + o (\|(h,k)\|).$$

If the derivative of $g$ exists with this definition, then $D_{\mathbf{u}} g (0,0) = Dg (0,0) \mathbf{u}$ [exercise], so that this definition of the derivative and the definition of the directionnal derivative are coherent. Note, however, that the converse is not true: the fact the directionnal derivatives exist do not imply that a ("full") derivative exists.

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I think it is a clear answer but i still got one question,why ppl would find $D_ug$ from $$D_ug(0,0) = \lim_{h\rightarrow0}\frac{g(hu,hv)-g(0,0)}{h}$$ rather than $$ \lim_{(h,k) \to (0,0))} \frac{g(h,k) - g(0,0) - Dg (0,0) \mathbf{u}}{\sqrt{h^2+k^2}} = 0,$$ as we all know,$$ \lim_{(h,k) \to (0,0))} \frac{g(h,k) - g(0,0) - Dg (0,0) \mathbf{u}}{\sqrt{h^2+k^2}} $$ not always equal to 0 In general . –  Mathematics Jan 15 '12 at 14:42
    
Sorry, I fear I do not understand this question. –  D. Thomine Jan 15 '12 at 15:41
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