What does $d\log\left(\frac{y}{x}\right)$ mean mathematically? I am used to seeing derivatives written as $$\frac{df}{dx}.$$
But my economics professor keeps using notation like $$ d\log\left(\frac{y}{x}\right)$$ and I have no idea what this means. What does this notation signify?
If it's a derivative, with respect to what? There is no denominator term. And I thought modern calculus used standard analysis with limits and not infinitesimals.
my question: 


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*What does  $d\log\left(\frac{y}{x}\right)$ signify?

*How do I take this derivative and with respect to what?

 A: If you have a function $f: \Bbb R^n \to \Bbb R$, say, $f(x_1,\cdots,x_n)$, then: $${\rm d}f = \frac{\partial f}{\partial x_1}{\rm d}x_1 + \cdots + \frac{\partial f}{\partial x_n}\,{\rm d}x_n. $$ In your case: $${\rm d}\left(\log\left(\frac{y}{x}\right)\right) = \frac{x}{y}\left(-\frac{y}{x^2}\right)\,{\rm d}x+ \frac{x}{y}\frac{1}{x} \,{\rm d}y = -\frac{1}{x}\,{\rm d}x + \frac{1}{y}\,{\rm d}y.$$
A: In economics, the log differential (called elasticity) represents the differential percent change 
$$d\log (x)=\frac{dx}{x}$$
of an economic variable.  For the elasticity of interest
$$d\log(y/x)=\frac{d(y/x)}{(y/x)}$$
which represents the percent change in the ratio $y/x$.  One can also express this as
$$d\log(y/x)=d\log (y)-d\log(x)=\frac{dy}{y}-\frac{dx}{x}$$
which represents the difference in differential percent changes.

If $U(x,y)$ is the utility function for consumption, then the elasticity of substitution is given by
$$\frac{d\log(y/x)}{d\log(U_x/U_y)}=\frac{\frac{d(y/x)}{(y/x)}}{\frac{d(U_x/U_y)}{(U_x/U_y)}}$$
where $U_x=\frac{\partial U}{\partial x}$, $U_y=\frac{\partial U}{\partial y}$, and $U_x/U_y$ is the marginal rate of substitution.
A: Are you familiar with the origin of the Leibniz notation, $\frac{{\rm d}y}{{\rm d}x}$? It comes from the derivation from first principles of the derivative being the limit of the gradient of secants of a function, where that gradient is calculated by measuring a change in $y$ relative to a change in $x$, i.e. $\frac{\delta y}{\delta x}$ as $\delta x \rightarrow 0$.
However, you can also look at something like $\frac{\rm d}{{\rm d}x}$ as being an operator which transforms a function to its derivative, so that $\frac{\rm d}{{\rm d}x} f(x) = \frac{{\rm d}f}{{\rm d}x}$.
And then we have integrals, which (at least if we're working with Riemann sums) are a limit of sums of areas of rectangles that approximate a function, as we make the width of the rectangles approach zero, i.e. $\int f(x) {\rm d}x$ is the limit of $\sum (f(x) \times \delta x)$ as $\delta x \rightarrow 0$.
So, in an incorrect but somewhat intuitive sense, ${\rm d}$ is somewhere between being "an infinitesimal change in something" and "an operator representing the basis of a derivative or integral".
So if you write ${\rm d}f(x, y)$, then you're trying to represent something about the behaviour of $f$, and in particular you can split that up as ${\rm d}f(x, y) = \frac{\partial f}{\partial x}{\rm d}x + \frac{\partial f}{\partial y}{\rm d}y$, which is a bit more meaningful in the context of either a derivative or integral - for example, $\frac{{\rm d}f}{{\rm d}x} = \frac{\partial f}{\partial x}\frac{{\rm d}x}{{\rm d}x} + \frac{\partial f}{\partial y}\frac{{\rm d}y}{{\rm d}x} = f_x+y'f_y$ or $\int {\rm d}f = \int \frac{\partial f}{\partial x}{\rm d}x + \int \frac{\partial f}{\partial y}{\rm d}y$.
A: Answers:


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We have a function f, ant a point x for that function. Very often we can represent change of function value near this point lineary with respect to change of argument "A*dx" and with some infitine small error "|dx|*o(dx)" if dx->0. So that linear change without "error" is called differenial and is df


2a. 
In your case df should be linear relative to dx. 
It is exist a theorem that if you have a function and has partial derivatives by all variable and all of that partial derivatives and continious then the function is differentiable, and you can represent it as f(x+dx)-f(x) =Adx+0(dx)|dx|
2b. 
It is exist one more theorem that if function is diffeenretiable then f(x+dx) -f(x) = gradient(f) * dx + 0(dx)*|dx|
p.s. The mathematic discipline which cover that 2-nd question is differentional counting on multivariable functions.
