I've been studying physics and I found this weird differentiation.

$\ln x = \ln a + \ln b$

Now differentiating both sides,

$\dfrac{dx}x = \dfrac{da}a + \dfrac{db}b$

First of all this weird differentiation doesn't make sense to me. I understand that $(\ln x)'$ will be $\frac1x$ and in no way $\frac{dx}x$.

So I asked a person about it and they replied with this:

Basically, they told me that we have differentiated both sides wrt $x$.

Now according to me, differentiating R.H.S. i.e. $\ln a$ wrt x should yield $0$. But according to them, it is

$\dfrac{d(\ln a)}{dx} = \dfrac{da}{adx}$

I don't get it!

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    $\begingroup$ Are $a$ and $b$ functions of $x$? Or are they constants? Or what? $\endgroup$
    – TonyK
    Commented Jul 4, 2020 at 9:45
  • $\begingroup$ I don't think they're functions of x. They're constants. I mean, they are variables, but it's physics so we replace them with values of a and b $\endgroup$ Commented Jul 4, 2020 at 12:32
  • 2
    $\begingroup$ Well then there's nothing to differentiate. $x=ab$ and that's that. $\endgroup$
    – TonyK
    Commented Jul 4, 2020 at 12:36
  • $\begingroup$ @TonyK take a look at a specific step on this page socratic.org/questions/… $\endgroup$ Commented Jul 4, 2020 at 12:39
  • 1
    $\begingroup$ @TonyK is right if a,b are constants then dx=0$ thats all. $\endgroup$ Commented Jul 4, 2020 at 12:45

3 Answers 3


$$x=ab$$ Differentiate with respect to $x$ $$1=a'b+b'a$$ $$dx=bda+adb$$ Since $x=ab$ we have: $$\dfrac {dx}{x}=\dfrac {da}{a}+\dfrac{db}{b}$$

If $x,a,b$ are function of $t$ then: $$x=ab$$ $$\dfrac {dx}{dt}=b\dfrac {da}{dt}+a\dfrac {db}{dt}$$ $$dx=bda+adb$$ Since $x=ab$: $$\dfrac {dx}{x}=\dfrac {da}{a}+\dfrac {db}{b}$$

  • $\begingroup$ I don't think they're functions of x... please take a look at a specific step on this page socratic.org/questions/… $\endgroup$ Commented Jul 4, 2020 at 12:33
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    $\begingroup$ If $a,b$ are constants then how comes you have $x=ab$ ? @Typedragon33 And if $a,b$ are constants then so is $x$ since $x=ab$ and the derivative of $x$ is then zero .. $\endgroup$ Commented Jul 4, 2020 at 12:35
  • $\begingroup$ I think I'm confused. I know they're not constants. But in the derivation, they are treated as variables (which makes sense). It was stupid of me to say they're constant. They are variables through and through. But I still don't get how they are functions of x? I'm confused $\endgroup$ Commented Jul 4, 2020 at 12:42
  • $\begingroup$ No problem. Differentiate $x=ab$ with respect to x and consider $a,b$ as functions of x...And look at what you get @Typedragon33 $\endgroup$ Commented Jul 4, 2020 at 12:48
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    $\begingroup$ hmm.. if I treat it like d(f(x)g*(x))/dx then I get d(ab)/dx = a(b)' + b (a)'. Guess they HAVE treated a and b as functions of x! Perhaps I got confused because I was looking at it from a physics perspective rather than a mathematical one. I guess.. $\endgroup$ Commented Jul 4, 2020 at 12:53

The differential of a differentiable two-variable function ($f(x,y)$)reads: $$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$$

In your case you have $x(a,b)=ab$, it has been used that $d(\log x)=\frac{dx}{x}$, $\frac{\partial x}{\partial a}=b=\frac{x}{a}$ and $\frac{\partial x}{\partial b}=a=\frac{x}{b}$


$\newcommand\diff{\mathop{}\!\mathrm{d}}$ Remark that we define $\big(\ln(u)\big)'=\dfrac{u'}{u}$

Applied to $\ln(x)$ and since $(x)'=1$ this gives $\dfrac 1x$.

The derivation can also be written $f'(x)=\dfrac{\diff f}{\diff x}(x)\quad$ or $\quad\diff f(x)=f'(x)\diff x$

The differential looks like an incomplete derivation where we would have omitted to divide by $\diff x$.

$\diff (\ln(u)\big)=\dfrac{\diff u}{u}$

Compared to the $u$ form at the beginning, we basically replaced the $u'$ by $\diff u$.

You use this when performing variable substitution in integrals, for instance let set $u=\sin(x)$

Then you do $\diff u=\cos(x)\diff x$ and go on replacing $\diff x$ in the integral.

Here you consider $a,b$ are new variables (not constants) and that $x$ is dependent of these.

  • $x=a+b$ then $\diff x=\diff a+\diff b$

  • $x=f(a,b)$ then $\diff x=\dfrac{\partial f}{\partial a}\diff a+\dfrac{\partial f}{\partial b}\diff b$

  • $f(x)=g(a,b)$ then $\dfrac{\partial f}{\partial x}\diff x=\dfrac{\partial g}{\partial a}\diff a+\dfrac{\partial g}{\partial b}\diff b$

So here it is $\dfrac 1x\diff x=\dfrac 1a\diff a+\dfrac 1b\diff b$

  • $\begingroup$ Thanks. Guess I'll just rote it. $\endgroup$ Commented Jul 4, 2020 at 12:38

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