Error analysis by differentiation 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!
 A: 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 $\frac{\partial x}{\partial b}=a=\frac{x}{b}$
A: $$x=ab$$
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 $x=ab$:
$$\dfrac {dx}{x}=\dfrac {da}{a}+\dfrac {db}{b}$$
A: $\newcommand\diff{\mathop{}\!\mathrm{d}}$
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 $\dfrac 1x\diff x=\dfrac 1a\diff a+\dfrac 1b\diff b$
