Problem in evaluating logarithm derivatives Given the property of the logarithm that $\log{xy} = \log{x} + \log{y}$, how would one take the 'derivative' of this?
To be more clear,
$\log{xy} = \log{x} + \log{y}$ (property of $\log$)
$D(\log{xy}) = D(\log{x} + \log{y})$ (i) (Take derivative on both sides)
Now, $D(\log{x} + \log{y}) = D(\log{x}) + D(\log{y})$ (ii) (Derivative of sum is sum of derivatives)
Combining (i) and (ii): $D(\log{xy}) = D(\log{x}) + D(\log{y})$ (iii)
Implies: $\frac{1}{xy} = \frac{1}{x} + \frac{1}{y}$ (Evaluate derivative of logarithm using $D(\log{x}) = \frac{1}{x}$
$\frac{1}{xy} = \frac{1}{x} + \frac{1}{y}$ looks false to me; e.g. while $\log{6}$ does equal $\log{2} + \log{3}$, $\frac{1}{6}$ does not equal $\frac{1}{2} + \frac{1}{3}$.
My first guess was that the issue was related to what variable I take the derivative with respect to, but I'd like to understand this a little more formally if someone could guide me.
What's going wrong in this example? 
 A: Yes, as you noted, it matters which variable you take a derivative with respect to. Since we have two variables, if we assume they are independent, then we need to use partial derivatives. If we chose $x$, then $$\frac{\partial \log(xy)}{\partial x}=\frac{y}{xy}=\frac1x$$and $$\frac{\partial}{\partial x}(\log x+\log y)=\frac1x+0=\frac1x$$So the answers agree. (the same thing happens if we chose $y$ instead).
Alternatively, you could take a total derivative. $$\mathrm d(\log xy)=\frac{\partial \log xy}{\partial x}\mathrm dx+\frac{\partial \log xy}{\partial y}\mathrm dy=\frac1x\mathrm dx+\frac1y \mathrm dy$$This agrees with $$d(\log x+\log y)=\frac1x\mathrm dx+\frac1y \mathrm dy$$
So there are no inconsistencies.
A: The issue is that your differential operator $D$ does not behave in the way you think it does!
$$
D(\log(xy))=\frac{1}{xy}D(xy)=\frac{1}{xy}(ydx+xdy)=\frac{1}{x}dx+\frac{1}{y}dy=D(\log(x))+D(\log(y))
$$
At no point in this computation do we actually have $\frac{1}{xy}=\frac{1}{x}+\frac{1}{y}$ -- we are working with differentials, and not derivatives.
