I was watching an online tutorial and saw this derivation.

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It seems the the author took the derivative with respect to y on left side and to x on right side. I thought dx should always be in the denominator and should on both side of the equation. Is it partial derivative? Or maybe my misunderstanding of the notation?

Could anyone explain how this works? FYI the link of the tutorial is https://www.youtube.com/watch?v=aXBFKKh54Es&list=PLwJRxp3blEvZyQBTTOMFRP_TDaSdly3gU&index=98, the differentials was taken at around 2'20"

Much appreciated! Happy New Year.

  • $\begingroup$ The second equation is rather the equality of the differentials (instead of the derivatives) of the functions in the first equation. $\endgroup$ – Pp.. Jan 1 '15 at 3:44
  • $\begingroup$ Maybe a notational distinction that people use is $\frac{\partial f}{\partial x}(x)$ (Leibnitz notation for the derivative) versus $\text{d}f=\frac{\partial f}{\partial x}(x)\text{d}x$ for the differential. $\endgroup$ – Pp.. Jan 1 '15 at 3:46
  • $\begingroup$ Perhaps you could also add a link to the online tutorial. (Both to add more context to your question and the link might be interesting for some people who stumble upon your post.) $\endgroup$ – Martin Sleziak Jan 1 '15 at 8:54
  • $\begingroup$ The author did not differentiate the LHS wrt $y$ and RHS wrt $x$. $\endgroup$ – whacka Jan 1 '15 at 9:54

From your other question I can say that it's a misunderstanding of differentiation and what is happening here. One simple way to understand it is as follows:

Take differentiable functions $x,y$ with open domain $D$ such that $\ln(y(t)) = a + b \ln(x(t))$ for any $t \in D$.

Then differentiating gives $\frac{y'(t)}{y(t)} = b \frac{x'(t)}{x(t)}$ for any $t \in D$.

If you rewrite this in Leibniz's suggestive notation you get:

$\frac{dy(t)}{y(t)\ dt} = b \frac{dx(t)}{x(t)\ dt}$

And if you treat $dt$ as a sufficiently small non-zero quantity and multiply both sides by it, you get:

$\frac{dy(t)}{y(t)} = b \frac{dx(t)}{x(t)}$

Note that this equation must be used with the understanding that the differentials are taken in the context of a small change in $t$, since we have now omitted $dt$.

Note that this is not equivalent to a small change in $x(t),y(t)$! For example if $x(t) = \sin(t)$ and $y(t) = \sin(2t)$ for any $t \in \mathbb{R}$, the curve $(x,y)$ intersects itself at $(0,0)$ and has two gradients there, one when $t = 0$ and the other when $t = \pi$.

And historically we have used bare variables to represent changing quantities so if we drop the parameter $t$ we get:

$\frac{dy}{y} = b \frac{dx}{x}$.

Note that this now makes no sense unless both differentials are taken in the same context, which now means that not only are they taken with respect to a small change in $t$ (which is now missing from the equation), we have to use this equation with the understanding that the values of $x,y$ are tied to each other, represented earlier on explicitly by the parameter $t$. In many cases in physics, however, $t$ is $x$ itself, which is expressed by the mathematically not quite right "$y$ is a function of $x$".

  • $\begingroup$ Thanks a lot. I posted the link of the video and the differentiation is between 2' to 2'20". Could you confirm that your explanation is consistent with the author's calculation? I just wanted to be sure about this. $\endgroup$ – nouveau Jan 1 '15 at 18:13
  • $\begingroup$ Well I don't have any background knowledge in that topic but I suppose that the author assumed that $y$ is dependent upon the '$x_k$'s and $x_1$ is independent of the other '$x_k$'s and he is only concerned about what happens to $y$ when $x_1$ changes. In that case, you can either use the multi-variable derivative to understand the whole thing, or you can just take my $t$ to be $x$ and treat the other '$x_k$'s as constants, which gives you the result. $\endgroup$ – user21820 Jan 2 '15 at 10:32

It's curious that you should call this a "derivation" because this is more accurate than saying that the derivative in the usual sense was taken. The construction used was the exterior derivative, or differential. This satisfies $$df(x,y)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$$ for a function depending only on $x$ and/or $y$, so if $f(x,y)=\ln x$, then $$df(x,y)=\frac{1}{x}dx+0dy=\frac{dx}{x}$$ and if $g(x,y)=\ln y$ then $$dg(x,y)=0dx+\frac{1}{y}dy=\frac{dy}{y}$$ The reason it is called a derivation is for any functions $p,q$ we have $$d(pq)=qdp+pdq$$

  • $\begingroup$ Thanks for replying. But I couldn't understand your post well. the df(x,y)=∂f∂xdx+∂f∂ydy is definitely the formula I should look into but i couldn't find it from the wiki page. is there a name for this formula? $\endgroup$ – nouveau Jan 1 '15 at 7:56
  • $\begingroup$ @nouveau: It's often called the total derivative. By the way you can see how the total derivative works via exactly the same understanding as in my answer. $\endgroup$ – user21820 Jan 1 '15 at 9:51
  • $\begingroup$ @nouveau I linked too quickly, here is a better link: en.wikipedia.org/wiki/Differential_of_a_function $\endgroup$ – Matt Samuel Jan 1 '15 at 10:30


$$ d( \ln y ) = \frac{1}{y} dy$$

$$ d( \alpha + \beta_1 \ln x_1 ) = \beta_1 \frac{1}{x_1} dx_1 $$

Since $\alpha, \beta_1 \in \mathbb{R} $

  • $\begingroup$ Thanks for replying. I am still confused. say there is f(x)=g(y), then we can get d[f(x)]/dx=d[g(y)]/dy? $\endgroup$ – nouveau Jan 1 '15 at 7:15
  • $\begingroup$ The above derivatives are with respect to x and y respectively. Without knowing the relationship between x and y, how can the derivatives be equal? Thanks. $\endgroup$ – nouveau Jan 1 '15 at 7:24

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