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Am I right that I can write/interpret any derivative $\frac{\partial f(x)}{\partial x}$ as derivative around zero, i.e.:

$\frac{\partial f(x)}{\partial x}=\left.\frac{\partial f(h+x)}{\partial h}\right|_{h=0}~?$

Is this interpretation/notation in any way common?

I am asking this basic question since I am trying to look at the Euclidean space as a special case of a (Lie) manifold.

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    Just apply the chain rule.2012-08-14

2 Answers 2

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Simply by using the definition of the derivative:

$\begin{align*} \frac{\partial f}{\partial x}(x) & = \lim_{y \to x} \frac{f(y)-f(x)}{y-x} \\ & \overset{y=x+\epsilon}{=} \lim_{\epsilon \to 0} \frac{f(x+\epsilon)-f(x)}{\epsilon} \\ & = \frac{\partial f(x+\epsilon)}{\partial \epsilon}(0) \end{align*}$

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    It is just a small subtly. I was think of the definition $\frac{\partial f(x)}{\partial x} := \lim_{\epsilon \to 0} \frac{f(x+\epsilon)-f(x)}{\epsilon}$, but not of the slightly more general one $\frac{\partial f(x)}{\partial x} := \lim_{y \to x} \frac{f(y)-f(x)}{y-x}$. Obviously, both are equivalent.2012-08-15
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This is correct.

Using the chain rule (thanks to nayrb) we get:

$\left.\frac{\partial f(h+x)}{\partial h}\right|_{h=0}=\left.\frac{\partial f(y)}{\partial y}\right|_{y=0+x}\cdot\frac{\partial (h+x)}{\partial h} = \frac{\partial f(0+x)}{\partial x}\cdot 1 = \frac{\partial f(x)}{\partial x}~.$