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# Is A-level Maths Wrong? watch

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1. Hi, guys I'm currently teaching my self multivariable calculus out of interest.
So far so good, until I noticed my entire understanding of maths from A-level and beyond broke down.

I'm currently learning about the chain rule lf multivarivable functions, however, I have noticed that dy/dx is no longer treated as a fraction as such so for example dx in the denominator can no longer be canceled out by multiplying it by dx/ds, example.

My question is, why is this condoned for single variable calculus but not multivariable calculus?
2. (Original post by An-Lushan)
...
It's not 'wrong' because it works. It's more efficient to use easier methods than harder ones. It's less rigorous, who cares?

Just like you'd write 1 + 1 = 2 without proving it from 1st principles, you take the less rigorous (easier) approach.

'Wrong' is overused and misunderstood, it's meaningless in the context of even slightly nuanced points.
3. (Original post by Physics Enemy)
It's not 'wrong' because it works. It's more efficient to use easier methods than harder ones. It's less rigorous, who cares?

Just like you'd write 1 + 1 = 2 without proving it from 1st principles, you take the less rigorous (easier) approach.

'Wrong' is overused and misunderstood, it's meaningless in the context of even slightly nuanced points.
Sorry if I wasn't clear, i meant the chain rule for partial derivatives not normal derivatives.Looking at the general form for this concept shows that derivatives are not actually fractions and never were.
The thing is I don't know why this is the case-we are just expected to accept it for what it is.
4. (Original post by An-Lushan)
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I know, I'm saying it's an unrigorous, visual trick when using the chain rule on normal derivatives (single variable) to make life easier. You must use the rigorous approach with partials (multi variable).

You should never have thought normal derivatives were actually fractions where num/denom cancels, was misunderstood or mistaught.
5. (Original post by Physics Enemy)
I know, I'm saying it's an unrigorous, visual trick when using the chain rule on normal derivatives (single variable) to make life easier. You must use the rigorous approach with partials (multi variable).

You should never have thought normal derivatives were actually fractions where num/denom cancels, was misunderstood or mistaught.
^^ I'm really no expert on maths but I must admit when I first saw this "cancelling" it really me off because it's an operation. I can sort of see why they do it, but he is right - calculus is a very picky and sensitive topic that can quite easily be spun off the wrong way. But get it right and it's massively powerful
6. You can treat them like fractions at a-level but my teacher constantly stressed that they aren't fractions which I mean totally makes sense because they aren't numbers.
7. I guess it's because the first derivative in single variable calculus represents a change in f(x)/change in x both of which are numbers so can be treated like fractions when cancelling things out.
8. (Original post by Physics Enemy)
I know, I'm saying it's an unrigorous, visual trick when using the chain rule on normal derivatives (single variable) to make life easier. You must use the rigorous approach with partials (multi variable).

You should never have thought normal derivatives were actually fractions where num/denom cancels, was misunderstood or mistaught.
Where can I find a convrete definition of what infitestimslly small is in calculus?

My textbook is pretty much saying:'You remember the chain rule from single variable calculus?Well, it applies here too.'

Even online many people that I've asked don't seem to understand it-they know it's not the correct mentality, but they simply accept it for what it is.

My guess is that infinitestimally small is not defined and could be interpreted as any extremely small number.
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Your 1st post says derivatives aren't numbers so can't be treated as fractions, 2nd post argues they are. The division operator by its defn applies to numbers and functions but not operators. So 3/5 and (x^2 + 1)/(e^x + 1) are fine, but (dy/dt)/(dx/dt) = dy/dx isn't, technically.
10. (Original post by An-Lushan)
I have noticed that dy/dx is no longer treated as a fraction as such so for example dx in the denominator can no longer be canceled out by multiplying it by dx/ds, example.

My question is, why is this condoned for single variable calculus but not multivariable calculus?
Can you put up a specific example? In fact, the "cancelling" approach works fine in at least some multi-variable cases, so it'd be nice to see the precise circumstances that are confusing you.
11. (Original post by atsruser)
Can you put up a specific example? In fact, the "cancelling" approach works fine in at least some multi-variable cases, so it'd be nice to see the precise circumstances that are confusing you.
Ok, the capital D reprrssenrs the partial derivativIt's difficukt to write out latex code on phone:
Z=f(x,y)

dz/dt=Dz/Dx(dx/dt)+Dz/Dy(dy/dt).

My intuition from A-level would tell me that
dz/dt=2Dz/dt
Again I know this is wrong because it just doesn't 'feel' right, but I'm having a hard time accepting it stemming from techniques used in single variable calculus at A-level.
12. (Original post by An-Lushan)
Ok, the capital D reprrssenrs the partial derivativIt's difficukt to write out latex code on phone:
Z=f(x,y)

dz/dt=Dz/Dx(dx/dt)+Dz/Dy(dy/dt).

My intuition from A-level would tell me that
dz/dt=2Dz/dt
Again I know this is wrong because it just doesn't 'feel' right, but I'm having a hard time accepting it stemming from techniques used in single variable calculus at A-level.
OK, I see what you mean. In fact, this is what I was thinking of, when I said that "cancelling" works fine with partial derivs.

However, you can't cancel the dy-s and dx-s here and add to get . You can see that, if we invent some notation that shows what we have after "cancelling" better:

After "cancelling", we can consider that we have a rate of change of z wrt t in each term, but it is a rate of change that depends on the two different variables, x and y. In general, these will not have the same values, so we can't simply forget about the intermediate x and y variables and write in both cases - they are not numerically (or algebraically) the same, so our notation must record that fact, which is precisely what the "uncancelled" version does anyway.

In short, if you cancel, then the "cancelling" works fine to indicate a rate of change of z with t, but we have two different rates of change, which your notation ignores.
13. (Original post by atsruser)
OK, I see what you mean. In fact, this is what I was thinking of, when I said that "cancelling" works fine with partial derivs.

However, you can't cancel the dy-s and dx-s here and add to get . You can see that, if we invent some notation that shows what we have after "cancelling" better:

After "cancelling", we can consider that we have a rate of change of z wrt t in each term, but it is a rate of change that depends on the two different variables, x and y. In general, these will not have the same values, so we can't simply forget about the intermediate x and y variables and write in both cases - they are not numerically (or algebraically) the same, so our notation must record that fact, which is precisely what the "uncancelled" version does anyway.

In short, if you cancel, then the "cancelling" works fine to indicate a rate of change of z with t, but we have two different rates of change, which your notation ignores.
Brilliant that clears everything up.Thanks for the help!

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