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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?

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  • $\begingroup$ Like the comment above said- you can’t take derivatives with respect to all different variables and expect them to be equal. $\endgroup$
    – user545426
    Commented Jan 5, 2019 at 4:19
  • $\begingroup$ (This comment was from me -- deleted when reading the fact that ayt the end of your question you acknowledged something was sketchy in the variables you differentiated with respect to. Essentially, I was saying you were "differentiating simultaneously wrt 3 variables," in different places: $xy$, $x$, $y$ -- and that's not allowed.) $\endgroup$
    – Clement C.
    Commented Jan 5, 2019 at 4:21
  • $\begingroup$ Thanks guys; this clears things up! I appreciate the fast responses. $\endgroup$ Commented Jan 5, 2019 at 4:24

2 Answers 2

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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.

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  • $\begingroup$ Thank you for the quick response; this is my favorite answer posted so far. $\endgroup$ Commented Jan 5, 2019 at 4:25
  • $\begingroup$ No problem, happy to help! $\endgroup$
    – John Doe
    Commented Jan 5, 2019 at 5:02
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    $\begingroup$ @user2192320 you may want to accept it then by clicking the tick mark on the left of the answer. $\endgroup$
    – Ruslan
    Commented Jan 5, 2019 at 16:24
  • $\begingroup$ Hey Ruslan, sorry for the late follow up; I accepted the response and checked the tick mark. Thanks for the guidance. $\endgroup$ Commented Jan 16, 2019 at 20:10
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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.

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