Problem with understanding a Differential in Multivariable Calculus I have just started with Partial Differentiation and the book from where I'm learning (Mathematical Methods in the Physical Sciences) had the following problem on approximations using differentials which I failed to understand:

The electrical resistance $R$ of a wire is given by $k\dfrac{l}{r^2}$ where $k$ is the constant of proportionality, $l$ is its $length,$ and $r$ is its $radius.$ If the relative error of length measurement is $5%$ and the relative error in radius measurement is $10%,$ find the largest possible relative error in $R.$ 

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The solution was given as follows:
$$$$Relative error in $l$ means the actual error in measuring $l$ divided by the length measured ie $\dfrac{dl}{l}$.$$$$
$$\Rightarrow \bigg|\dfrac{dl}{l}\bigg|=0.05$$
$$\Rightarrow \bigg|\dfrac{dr}{r}\bigg|=0.1$$
We can find $\dfrac{dR}{R}$ by differentiating $\ln(R)$. From $R=k\dfrac{l}{r^2},$
$$\ln(R)=\ln(k)+\ln(l)-2\ln(r)$$ Then $$\dfrac{dR}{R}=\dfrac{dl}{l}-2\dfrac{dr}{r}$$
This is where I am confused. In all the previous problems, we had to use the fact that $$dz=\dfrac{\partial}{\partial x} dx+\dfrac{\partial}{\partial y}dy$$ is a good approximation to $\Delta z.$$$$$However, using this in the $'\ln'$ equation, I get:
$$d(\ln(R))=\dfrac{1}{l}\times dl-2\times\dfrac{1}{r}\times dr$$
How can we say that $d(\ln(R))=\dfrac{dR}{R}?$ I know that we can write $dy(x)=\dfrac{d}{dx} y(x)\times dx$ But $R$ is a function of $2$ independent variables $l,r.$ Since $R$ is not an independent variable like $x$, how can we differentiate directly with respect to $R?$ $$$$ In One Variable Calculus, I had learnt that for example $$\dfrac{d}{d \cos (x)} e^x=\dfrac{\frac{d}{dx} e^x}{\frac{d}{dx} \cos(x)}$$ since $\cos (x)$ is not an independent variable; it is a function of $x$
$$$$Similarly, $R$ is a function of the independent variables $l,r.$ How can we then write $d(\ln(R))=\dfrac{dR}{R}?$
$$$$I am extremely confused with this and have been thinking about this since quite a few days. I cannot express how grateful I would be if somebody please cleared this doubt of mine. Many, many, many thanks in advance!
 A: You have $$R=k\frac{l}{r^2}$$ Taknig logarithms of both sides $$\log(R)=\log(k)+\log(l)-2\log(r)$$ Differentiating with respect to $l$ gives $$\frac 1R \frac{dR}{dl}=\frac 1l$$ that is to say  $$\frac{d R}R=\frac{d l}l\to \frac{\Delta R}R=\frac{\Delta l}l$$Similarly$$\frac 1R \frac{dR}{dr}=-\frac 2r$$ that is to say  $$\frac{dR}R=-2\frac{dr}r\to\frac{\Delta R}R=-2\frac{\Delta r}r$$Since you look for the total error (individual errors are not signed) $$\bigg|\frac{\Delta R}R\bigg|=\bigg|\frac{\Delta l}l\bigg|+2\bigg|\frac{\Delta r}r\bigg|$$
A: By $d(\ln(R)) = \frac{dR}{R}$ you are differentiating the function $f : x \, \mapsto \ln(x)$ which is a univariate function. You are not differentiating the function $R$.
Just think of that as differentiating a composition of functions:
$$d(\ln(R)) = d(f \circ R)(l,r) = df(R(l,r))\cdot dR(l,r) $$
If you continue the differentiation of $d(\ln(R))$ you will get:
$$d(\ln(R)) = \frac{dR}{R} = \frac{1}{R}\left( \frac{\partial R}{\partial l} dl + \frac{\partial R}{\partial r} dr\right)$$
And after some simplification you will find that the RHS just equals to
$$ \frac{dl}{l} - 2\frac{dr}{r}  $$
