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The origin of dy = f'(c)dx in differentials

by vanmaiden
Tags: differentials, equation, origin
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vanmaiden
#1
Nov21-11, 10:33 PM
P: 101
1. The problem statement, all variables and given/known data
If you have a point at x = c and a function f(x), then I know Δy = f(c + Δx) - f(c).
Also, dy = f'(c)dx. However, I am uncertain of the origin of dy = f'(x)dx.
I want to say:
f(c + Δx) - f(c) = f'(c)(x-c) was simplified to dy = f'(c)dx
where f(c + Δx) - f(c) = dy
(x-c) = dx
But that would result in Δy = f'(c)dx and Δy > dy. Hence, it is...or should be wrong. Could anybody clairify the origin of dy = f'(c)dx for me?

2. Relevant equations
Im the problem, actually.


3. The attempt at a solution
Also, in the problem! lol
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CompuChip
#2
Nov22-11, 01:03 AM
Sci Advisor
HW Helper
P: 4,300
Actually, the trick is to look at the ratio of the changes, [itex]\frac{\Delta y}{\Delta x}[/itex]. You can represent this on the graph of the curve as the average change of the function, or the slope of the straight line approximation on that small interval.
Initially it may seem strange to use a straight line approximation, but if you think about it for a minute, you will see that it's not such a bad idea because the smaller you make [itex]\Delta x[/itex], the better the straight line resembles the graph around x = c. In fact, the line you get becomes closer and closer to the tangent line of the graph at x = c. So it makes sense to try and do this analytically, for which we can use a limit: we consider
[tex]\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}[/tex]
And to make the dependence of the numerator explicit,
[tex]\lim_{h \to 0} \frac{f(c + h) - f(c)}{h}[/tex]
where I called [itex]\Delta x = h[/itex].

This number, if the limit exists, actually is the slope of the tangent line. Note that is no longer a real quotient, it's just the limit of a quotient. So we can invent the notation [itex]\frac{dy}{dx}[/itex] or [itex]\frac{df(x)}{dx}[/itex] or, if you want to have c in there somewhere,
[tex]\left. \frac{df(x)}{dx} \right|_{x = c}[/tex]
The safest way is now to think of it as just that, notation, and nothing else. Again, df/dx is not really a fraction anymore, so if df/dx = 3 that merely means that the slope of f at x is equal to 3. You shouldn't allow yourself to write things like df = 3 dx. If you can't resist the temptation, then consider writing f'(c) instead :-)

Just as an after-thought, it is true (for continuous functions) that if [itex]\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = f'(c)[/itex], then
[tex]\lim_{\Delta x \to 0} \Delta y = \lim_{\Delta x \to 0} f'(c) \Delta x[/tex]
because f'(c) is just a fixed number (which I denote by f'(c), or you could call it a or L) and if you plug in [itex]\Delta y = f(c) - f(c + \Delta x)[/itex] you will find that both sides of the expression are equal to 0. So in that sense, you can split out the fraction, but it doesn't give you anything useful.
Bacle2
#3
Nov22-11, 01:11 AM
Sci Advisor
P: 1,170
Basically, you set y=f(x); then, Δy=f(c+Δx)-f(c) is the actual change in the value of f from
c+Δx to c. If f is differentiable at c, then the change in values of f can be approximated at any level of precision (in a strict δ-ε sense) by a linear function . This approximation is called dy, and is given by dy=f'(x)dx (or a related expression in higher dimensions), so that, limitΔx→0Δy=dy , i.e., the change in the values of a differentiable function can be modeled with indefinite precision by a linear function; this is actually too, the actual definition of differentiability: f is differentiable if the change of values of f can be modeled locally (in a 'hood of a point where f is differentiable.) by a linear function L; this L is called the differential.


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