# calculus stuff,Watch

This discussion is closed.
#1
I just realised I don't really understand what some of these things mean. for example, we are just told to wtite dy/dx. I'm guessing this means something like "the change in y over the change in x, as the change in x tends to zero" or something.

but then we are told to write as an integral "dx" after the thing to be integrated, and that this means "with respect to x". which doesn't make a lot of sense.

Then a day or two ago I saw people doing things like:

dy/dx = 5x
dy = 5x dx

and things like that. is this allowed? can you multiply through by dx?
what do dx and dy mean?

With things like implicit differentiation and integration by parts still to be learnt I think it would help that I actually understood what I was doing with thr notation.

thanks.
0
14 years ago
#2
yes because with implicits you can

if youre differentiating x wrt y

dx/dy = 1/(dy/dx)

if thats the stuff you meant.

but then we are told to write as an integral "dx" after the thing to be integrated, and that this means "with respect to x". which doesn't make a lot of sense.
why not, cause that#s what you're doing

ie int(2x + 1)dx = x^2 + x +c
youre dealing with xs.

int(2t + 1)dt = t^2 + t + c
dealing with ts..
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14 years ago
#3
the integral dx probably means something like
\$^(a) _(b) f(x) dx [thats the integral from a to b of f(x)]

sum f(x) according to the change in x (ie the limits)

thats why you cant integrate if you have

\$^(u = a) _(u = b) f(x) dx

theres no change in the limits for any integration to be performed; no link between x and u... see?

techically i dont think you can multiply thru by dx but.. im not sure what the precise reason for how it all works is.

with implicit, you cant differentiatiate wrt the wrong variable so

d/dx(y) becomes d/dy(y) * dy/dx (by the chain rule)

we can differentiate the first part; its just 1

so this just becomes dy/dx (we just cant go any further)
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#4
yes, but dy/dx doesn't mean "with respect to y over with respect to x"
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14 years ago
#5
(Original post by mik1a)
yes, but dy/dx doesn't mean "with respect to y over with respect to x"
it means diff y wrt x.
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14 years ago
#6
i was in the same situation last year, i didn't know what the teacher was going on about when he said with respect to. I though respect means a totally different thing lol.

This year i worked it out by reading around the topic.
dy/dx=5x
dy = 5x dx

it's simple algebra, nothing complicated in it. It just looks complex.

dy/dx is just the gradient of a point in the curve.
d^2y/dx^2 is just differentiating twice.
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14 years ago
#7
(Original post by nas7232)
This year i worked it out by reading around the topic.
dy/dx=5x
dy = 5x dx

it's simple algebra, nothing complicated in it. It just looks complex.
i think mik1a is asking how you're allowed to do it though. I mean "dee y by de ex" is just a notation as is f'(x) and f''(x) ??

also why when you integrate you get the area under the curve
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14 years ago
#8
Oh, it just means differentiating y with respect to x. if it was other way around i think it would be 1 over it. ocr MEI p2 explains it thoroughly i think.
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#9
(Original post by lame)
it means diff y wrt x.
Yes I know we are told that
why does it not mean what I said?
they don't work together.
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14 years ago
#10
(Original post by mik1a)
Yes I know we are told that
why does it not mean what I said?
they don't work together.
dy/dx means

the change in y over the change in x

(delta means 'the change in')

think of differentiation from first principles (taking loads of gradients) and youll see what it means...
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#11
(Original post by nas7232)
i was in the same situation last year, i didn't know what the teacher was going on about when he said with respect to. I though respect means a totally different thing lol.

This year i worked it out by reading around the topic.
dy/dx=5x
dy = 5x dx

it's simple algebra, nothing complicated in it. It just looks complex.

dy/dx is just the gradient of a point in the curve.
d^2y/dx^2 is just differentiating twice.
but we are taught these two rules.

y = f(x)
dy/dx = f'(x)
from what you say we can write:
dy = f'(x) dx

but we are also taught that if

dy/dx = f'(x)
y = INT f'(x) dx

do the two things in bold have the same meaning?
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#12
(Original post by kikzen)
dy/dx means

the change in y over the change in x

(delta means 'the change in')

think of differentiation from first principles (taking loads of gradients) and youll see what it means...
Yes I understand that part.
But when you consider we are taught that in an integral, dx means "with repsect to x", this is not consistent notation!
0
14 years ago
#13
(Original post by mik1a)
but we are taught these two rules.

y = f(x)
dy/dx = f'(x)
from what you say we can write:
dy = f'(x) dx

but we are also taught that if

dy/dx = f'(x)
y = INT f'(x) dx

do the two things in bold have the same meaning?
well in one you have dy = and in another you have y = so no, theyre not teh same.

hmm i think youre just thinking too hard about this all. i think its all about usage; just like you can have words that mean different things in different situtations so, this is the same
0
14 years ago
#14
(Original post by kikzen)
well in one you have dy = and in another you have y = so no, theyre not teh same.

hmm i think youre just thinking too hard about this all. i think its all about usage; just like you can have words that mean different things in different situtations so, this is the same
Funnily enough they are the same - its called the fundamental theorem of calculus.

BUT it is mathematically illegal to say:

dy/dx = f'(x)
=> dy = f'(x) dx

although it those work out.

Euclid
0
14 years ago
#15
its funny how everyone first takes the dy/dx as gospel and then when you see like dx/dt you flip out because you don't know what this means. calculus is a process of self discovery haha
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14 years ago
#16
dy/dx = y/x hehe
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#17
lol yeah cancel the d
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14 years ago
#18
(Original post by Euclid)
Funnily enough they are the same - its called the fundamental theorem of calculus.

BUT it is mathematically illegal to say:

dy/dx = f'(x)
=> dy = f'(x) dx

although it those work out.

Euclid
Not really, we do it all the time, we integrate both sides:

Int 1*dy = Int f'(x)*dx

If we integrate 1 with respect to y we get y, so it follows that

y = f(x) + c

The statement dy = f'(x)dx is saying that an infinitesimal change in dy will produce an infinitesimal change in dx such that the ratio between these changes is f'(x). And there is nothing mathematically incorrect about doing so.

But when you learn partial differentiation then you must realise that you cannot do the above, the reason you can do it with the ones above is because they are total differentials. There are equivalent statements like the one we have looked at for functions of many variables, but they are matrix equations and have an object called the Jacobian instead of the "gradient".

Also there are formulations of calculus that treat the object "d" as an operator in it's own right, the differential, and then it would be perfectly valid to talk about dy or dx in terms of each other. So it depends on your notation really.
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14 years ago
#19
what does f'(x) mean?

does it mean dy/dx of f(x)
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14 years ago
#20
(Original post by nas7232)
what does f'(x) mean?

does it mean dy/dx of f(x)
yes

and d^2y/dx^2 = f''(x)
0
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