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How do you take the derivative of this?

k and r are constants

$E=kq\hat{r}(\frac{1}{r^{2}})$

$\frac{d}{dq}(E)=\frac{d}{dq}(kq\hat{r}(\frac{1}{r^{2}}))$

$\frac{d}{dq}(E)=(\frac{k\hat{r}}{r^{2}})\frac{d}{dq}(q)$

This is where I am now. And the answer is posted below.

$dE=(\frac{k\hat{r}}{r^{2}})dq$

How did they get this? Did they multiply both sides by dq?

Hack Delta
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  • Welcome to M.SE... and to physicist life!!! The expression you're asking for is known as a differential of a fucntion. It has a nice (and deep) mathematical expression, bur for the momment you can think about it just as very very (infinitesimal to be able to integrate it) ratio. The idea is the following: Suppose we have a electric field generate by a charge $q$, and that we add a very little charge $dq$. How changes te electric field? Well, since $dq$ is very small, I can assume that the change will be little. How little? Here it comes the interpretation of the derivative as a rate of change. – Dog_69 Mar 07 '19 at 23:17
  • My charge $dq $ is so small that I can approximate $dE(q+dq)$ by the tangent line using the derivative at that point. Or if you prefer, that the changes $dE$ and $dq$ are so small than I can approcimate their quotient (as a quotient of small numbers) by the actual derivative $dE/dq$ or, if you prefer, $E'$, in order to distinguish from the actual quotient. The idea is basically this, even though mathematics behind are much more difficult (and interesant) and the majority of physics don't care about it, unfortunately. – Dog_69 Mar 07 '19 at 23:20

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Yes they did, but of course multiplying by the infinitesimal $dq$ that is not a valid operation. The way to think about it is that this is shorthand for saying that if the derivative is $\frac{k \hat{r}}{r^2}$ then for sufficiently small variations in $q$, the change in $E$ is arbitrarily close to $$ \delta E = \frac{k \hat{r}}{r^2} \delta q $$ You don't get into trouble with this sloppy notational convenience until you encounter functions with discontinuities or singularities. But it is worth think about the issue each time, because when you do encounter a case where the shortcut notation fails, the subtle error resulting is going to be a massive headache.

Mark Fischler
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