Is dy/dx a Limit or a Quotient?

[Ashu2912]

dy/dx - a limit or a quotient??

Hello Friends. I have a confusion in differentiation. We know that
lim $$\Delta$$x->0 ; $$\Delta$$y/$$\Delta$$x = dy/dx.
Then how can dy and dx be treated as independent variables, and dy/dx as their quotient, if dy/dy is actually a limit?
Also, since we won't be able to write the above limit as
lim $$\Delta$$x -> 0; $$\Delta$$y / lim $$\Delta$$ x -> 0; $$\Delta$$x
as lim $$\Delta$$x -> 0; $$\Delta$$ x = 0, what does the independent dy and dx represent?

 

[statdad]

"Then how can dy and dx be treated as independent variables, and dy/dx as their quotient, if dy/dy is actually a limit?"

The use of $\frac {\mathroman{d}y}{\mathroman{d}x}$ as a fraction is simply a memory device and an aid to notation.

 

[HallsofIvy]

First, let me point out that Newton thought of it as a quotient- the ratio of "infinitesmals" but was not able to put "infinitesmals" on a rigorous basis so that limits took over mathematics. Within the last 50 years, "non-standard analysis" has put infinitesmals on a rigorous basis (with some very deep concepts from logic) so that we can, in that sense, think of the derivative as a fraction.

However, even without "infinitesmals" you can show that the derivative always "acts" like a fraction- for any fraction property, go back before the limit, use the fraction property of the "difference quotient" then take the limit again. So that symbolic separation of dy/dx into "dy" and "dx" is valid and very useful. Also, the concept of the "differential" becomes especially important in differential geometry.

 

[dextercioby]

$\frac{dy}{dx}$ is just a notation, it's not a real fraction, so 'dx' is not really in the denominator and cannot be be multiplied/simplified through as one simulates in separable ODE's of first order

$$\frac{dy}{dx} = f(y)g(x)$$ (1)

So normally (1) one should be written

$$y'(x) = f(y) g(x)$$ (1')

In geometric terms (1') is an equality of 2 0-forms. What one can do now in (1') is to multiply both terms of the equality (1') by the R-1-form dx (exterior product of 0-forms and 1-forms is equal to a 1-form, 0-forms are <scalars> wrt the exterior product) and get

$$y'(x) dx = f(y) g(x) dx$$ (2)

But the LHS in (2) is nothing but the exterior differential of the 0-form y, the 1-form dy.

$$dy = f(y) g(x) dx$$ (2')

Then

$$\frac{dy}{f(y)} = g(x) dx$$ (2'')

is again an equality between 1-forms. Integrating on a path in the (x,y) plane one find the solution to the initial ODE in an implicit form:

$$F(y) = H(x)$$