Question

What is a differential? I get that

d/dx f(x) = lim ∆x -> 0 (f(x + ∆x) - f(x))/∆x. But that to me seems as if the differential of x, dx, is some theoretical constant that is 0.0000....01 (infinite 0s). But I have heard many times that this isn't the case. So what is the meaning of the notation dy/dx ?

Answer 1

http://en.wikipedia.org/wiki/Differential_of_a_function

Answer 2

Forgive me if I am asking obvious questions, but I still dont understand the true nature of dx. How is dx varying if it is another real variable just like x?

Answer 3

\[\frac{dy}{dx}\] simply means how much your y change \[dy\] if you change x in the infinitesimally small value\[limit\ \Delta x\rightarrow 0\]. It's same by saying how big/stiff is the gradient/slope of a line if we vary the X by very very small amount (Delta x) from that X point.

Answer 4

I am just confused at what this infinitely small change in x actually is, or does its value theoretically change with every function?

Answer 5

I guess what I am asking is this: Is dx a constant or a variable?

Answer 6

and why?

Answer 7

@wikiemol , prob the graphical explanation will help you more ince i have limited space and time here, I'll just give u this khan academic link explaining about derivative. i hope it can clear everything up. https://www.youtube.com/watch?v=ANyVpMS3HL4

Answer 8

well yeah I understand all of that, its just it seems like the definition of a derivative where dy/dx = lim ∆x -> 0 (f(x + ∆x) - f(x))/∆x defines dy and dx together, but not individually. in other words, when you are dealing with changes in variables that aren't infinitely small, you can define a change in x seperately from a change in y. and so it seems as if the definition of a derivative is saying that dy = lim ∆x -> 0 (f(x + ∆x) - f(x)) and dx = lim ∆x -> 0 ∆x but that doesn't really make sense or hold any meaning. so when we make statements such as dy = f'(x) dx, it seems as if we are also saying lim ∆x -> 0 (f(x + ∆x) - f(x)) = f'(x) (lim ∆x -> 0 ∆x) but that doesn't make sense to me.

Answer 9

@wikiemol you are correct in understanding.
Leibnitz introduced this simplified notation of dy/dx
\[f'(x)={d\over dx}[f(x)]=\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}
\]

Answer 10

but what I am saying is, if we are defining dy/dx in this manor, how are we able to manipulate dy and dx as if they have been defined separately.

Answer 11

It seems that we have defined the derivative of a function without defining the differential of a variable.

Answer 12

because,
dx means a change in "x" (infinitesimal but still a number)

Answer 13

dx is nothing but \(\Delta x\) but very very small, indistinguishably small

Answer 14

See
http://en.wikipedia.org/wiki/Differential_of_a_function#Definition
for a discussion

Answer 15

so ∆x is a variable and dx is some arbitrarily small number so that ∆x ~ dx?

Answer 16

dx is \(\Delta x\) when it is almost zero (i.e., infinitesimally small).
In Taylor Expansion series, you use \(\Delta x\) as finite and not infinitesimally small. and hence you need infinite derivatives to approximate that small-ness