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I was always taught that $\frac d {dx} (\ln x) = \frac 1 x$. No derivative had as a result any $dx$ words. In a physics book I encountered something like this (error discussion) [there might be a little difference, as I don't have the book right now]:

We have $\ln a + \ln b = \ln c$. So after deriving both sides we get $\frac {da} a + \frac {db}b = \frac {dc}c $

Why is this small difference $da, db, dc$ put in there? Why isn't it $\frac 1a + \frac 1b = \frac 1c $ ? And what are we differentiating in reference to? $x$, $a$ or what?

Another example [this is exactly like this]: We have $[(\frac 1 {S_1}+\frac 1 {S_2})\rho g V + p_0]V = nRT$. Let's treat V as a function of T and differentiate both sides wrt T. Then we have

$2(\frac 1 {S_1}+\frac 1 {S_2})\rho g V \frac {dV} {dT} + p_0 \frac {dV} {dT} = nR$

Once again, why are we adding this difference quotient $\frac {dV} {dT}$? Is this as the derivative of the internal function?

Alfred Centauri
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marmistrz
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3 Answers3

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Regarding your first question, note that

$$\frac{d}{dx}\ln c = \frac{1}{c}\frac{dc}{dx}$$

Thus

$$\frac{1}{a}\frac{da}{dx} + \frac{1}{b}\frac{db}{dx} = \frac{1}{c}\frac{dc}{dx}$$

Then, as physicists often do, 'cancel' the $dx$ on both sides to yield

$$\frac{da}{a} + \frac{db}{b} = \frac{dc}{c}$$

Regarding your second question, it is stated that $V = V(t)$ thus it is elementary that

$$\frac{d}{dt}KV(t) = K\frac{dV}{dt}$$

Alfred Centauri
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  • Compare the comment on the Jim's answer - why can we cancel out the $dx$ when the derivative (the first equation is clearly a derivative of a compound function) is not a ratio? – marmistrz Dec 21 '14 at 14:49
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    @marmistrz, from the Taylor series, we know that $$f(x + dx) = f(x) + f'(x)dx$$ So, formally, for the log function, this is $$\ln(x + dx) = \ln(x) + \frac{1}{x}dx$$ or $$\ln(x + dx) - \ln(x) = d, \ln(x) = \frac{1}{x}dx$$ – Alfred Centauri Dec 21 '14 at 14:59
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    @marmistrz in physics, derivatives are viewed as ratios of small changes in the limit as the size of the change goes to zero. The functions we work with are generally well-behaved enough that this is valid. – David Z Dec 21 '14 at 16:35
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    You may prefer to think of this as simply a shorthand notation, in which $da$ really means $da/dx$. It's just another step onwards from writing what should strictly be $c(x)$ as $c$. – Harry Johnston Dec 22 '14 at 03:11
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The reason you see the $da$, $db$, or $dc$ is precisely because it isn't specified what the derivative is taken with respect to. What happens is the following:

First, they write the differential element of the logarithmic function:

$$d\,\ln f=\frac{df}{f}$$

This is always true no matter what function $f$ is. When we take a derivative with respect to some parameter $q$, we get:

$$\frac{d\,\ln f(q)}{dq}=\frac{1}{f(q)}\frac{df(q)}{dq}$$

So if $q=x$ and $f(q)=q$ then this becomes:

$$\frac{d\ln x}{dx}=\frac{1}{x}\frac{dx}{dx}=\frac{1}{x}$$

For your second example, the original equation was differentiated wrt $T$. Since, apparently in this case, only $V$ and $T$ itself were dependent on $T$, they are the only terms that get differentiated. $\frac{dT}{dT}=1$ and $\frac{dV^2}{dT}=2V\frac{dV}{dT}$. That's about it

Jim
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  • But I was taught that the derivative itself is not a ratio, although it graphically looks like a ratio. So why can we multiply by $df$ in the first equation? – marmistrz Dec 21 '14 at 14:48
  • @marmistrz because that is a differential element, not a complete derivative. It is a way of representing a derivative. Since it looks like a ratio, we can write it this way to mean that this is a derivative where we have not yet specified what it is taken wrt yet. Alternatively, it also can be used for integration in this form – Jim Dec 21 '14 at 14:53
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Using $V=V(T)$, taking the other terms as constant and working backwards from the result you gave, I suspect your equation of state actually reads $$ nRT = \left[ \left( \frac 1{S_1} + \frac 1{S_2} \right)\rho gV + p_0\right]\cdot V = \left( \frac 1{S_1} + \frac 1{S_2} \right)\rho gV^2 + p_0V $$ In that case, $dV/dT$ is just notation for the derivative and you get to your result by using the chain rule, specifically $$ \frac {d}{dT}V^2 = (V^2)' = 2VV' = 2V\frac{dV}{dT} $$ You can't quite interpret the other example that way.

One way to look at it is to just think of $\mathrm d$ as the 'infinitesimal' version of $\Delta$. In particular, if $\Delta a$ is small enough, we have $$ \Delta\ln a = \ln(a+\Delta a)-\ln a = \frac{\Delta a}a+\text{something very small} $$ and symbolically for $\Delta a\to0$ $$ \mathrm d\ln a = \frac{\mathrm da}a $$

A second way to look at your example is to regard $a,b,c$ as functions of a parameter $\lambda$. Then, differentiating $$ \ln a + \ln b = \ln c $$ with regards to that parameter yields (again by chain rule) $$ \frac {a'}a + \frac {b'}b = \frac {c'}c $$ The third way to look at this expression comes from differential geometry and requires the notion of differential forms.

The first step on the way is to realize that derivatives are actually linear maps, so for a function $f$, its derivative $f'$ should actually be considered a $1\times1$ matrix.

You can expand a matrix in terms of its basis and dual basis, ie $$ (a_{ij}) = \sum_{ij} a_{ij} \mathbf e_i\otimes \mathbf e_j^* $$

For functions $f:\mathbb R\to\mathbb R$, the target space in 1-dimensional and we can drop the sum over $i$ as well as the basis vectors $\mathbf e_i$. Then, for reasons I won't get into here, we rename $\mathbf e_1^*$ to $\mathrm dx$ and arrive at $$ \mathrm df = (f') = f'\mathrm dx $$ where the second term is said $1\times1$ matrix.

Christoph
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