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mlt

df/dx

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Lets assume we have the following function: f(x) = x^2 then: f'(x) = 2x df/dx = f'(x) We can then write: df = f'(x) dx but what is dx? Is it just (x0+h)-x0?

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Sort of. That's what it is if you want to evaluate the slope at an instantaneous location.


Rise Slope Run
(f0+h)-f0 = f'(x) * ((x0+h)-x0)


You just have to remember that the Rise/Run you get will always be equal to the tangent at x no matter what you plug in for h. For example, if you have a parabola but set x = the point at the bottom of the parabola, (f0+h)-f0 will always be zero no matter what you set h to.

h is effectively supposed to be a 'limit that approaches zero' for this reason. (and so are df and dx - they are both numbers that are approaching zero at a ratio defined by f'(x))

People usually do not treat df or dx as meaningful values; they use them as extra notation to make sure their calculus formulas do what they expect...

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Quote:
Original post by mlt
Lets assume we have the following function:

f(x) = x^2

then:

f'(x) = 2x
df/dx = f'(x)

We can then write:

df = f'(x) dx

but what is dx? Is it just (x0+h)-x0?


Actually, the "df/dx" is a formal symbol which stands for the function f', it is not the quotient of two variables "df" and "dx". So using the tools you have developed in calculus, it is not valid to write "df = f'(x) dx" for all you have defined is a function called "df/dx".

Strictly speaking you should write:

df/dx = f' (this is function equality)

or

(df/dx)(x) = f'(x) for every x in the domain of f' (i.e. the functions agree at every point, which is in the definition of function equality)

As alvaro mentioned, it is possible to make "df" or "dx" formal symbols too by dy defining them as linear forms. This is usually done in a second class in undergraduate analysis.

But the expression "df = f'(x) dx" is often used without having formally defined "df" or "dx", in contexts where it doesn't matter. In fact, you can do that and most intuitive manipulations you will do will turn out to produce a correct solution when you want to solve differential equations for example. You can also see it as a remainder that the derivative f' is a linear approximation of f around x, so that you can write:

Δf = f'(x) Δx

And Δf will be a good approximation of f(x + Δx) for sufficiently small Δx.

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Quote:
Original post by alvaro
Actually, dx is a 1-form. But that's something you can ignore for the rest of your life and nothing bad will happen to you.


... until you realize how cool general relativity is, then the 1-form becomes your best friend!! :D

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