# Derivative as derivative around zero?

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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I would say that you are right. – enzotib Aug 14 '12 at 13:50
Just apply the chain rule. – nayrb Aug 14 '12 at 13:51

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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Thanks, that's what I was looking for. I was not aware of the definition in the first line. However, a '=' is missing in the first line. I tried to change it, but it say "Edits must be at least 6 characters; is there something else to improve in this post?". Can you edit it? – Hauke Strasdat Aug 14 '12 at 21:22
Done. But, erm, how exactly did you define the derivative? – Najib Idrissi Aug 15 '12 at 7:27
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. – Hauke Strasdat Aug 15 '12 at 9:50

This is correct.

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

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

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