proof of dy/dx *dx/dy=1

MEPS1996

is it acceptable to prove dy/dx * dx/dy=1 in the same way as the chain rule is proved, ie like this:

1= deltay/deltax * deltax/deltay

where delta represents the greek letter delta reperesenting a small but finite change in the quantity

take limits of both sides as deltax goes to 0

1=dy/dx * dx/dy

is this an acceptable proof?

Reply 1

krisshP

14

Let's say we have a straight line where an increase of x by 2 gives an increase in y by 5. Then

dy/dx=change in y/change in x
=5/2

dx/dy=change in x/change in y
=2/5

So, dy/dx * dx/dy =(5/2)*(2/5)=1, just a case of cancelling out.

For a general case,

dy/dx * dx/dy
=(change in y/change in x) * (change in x/change in y)
=1, just by cancelling.

Reply 2

user2020user

16

It's not really a rigorous one. I think there's something to do with the Inverse Function Theorem that'll help you out.

Reply 3

Smaug123

15

Original post by MEPS1996

is it acceptable to prove dy/dx * dx/dy=1 in the same way as the chain rule is proved, ie like this:

1= deltay/deltax * deltax/deltay

where delta represents the greek letter delta reperesenting a small but finite change in the quantity

take limits of both sides as deltax goes to 0

1=dy/dx * dx/dy

is this an acceptable proof?

That gets a bit confusing when you come to things like

\dfrac{\partial y}{\partial x}\dfrac{\partial z}{\partial y}\dfrac{\partial x}{\partial z}=-1

.

The better proof goes like:

1 = \dfrac{d}{dy}(y) = \dfrac{d}{dy}(f(f^{-1}(y))) = \dfrac{d (f^{-1}(y))}{dy} (f'(f^{-1}(y)))

by chain rule.
If we now let

f^{-1}(y) = x

, we get

1 = \dfrac{dx}{dy} f'(x)

, which is what we wanted.

(I think that's rigorous in the case of

f

being sufficiently nice - it's enough for it to be continuously differentiable and bijective with everywhere-nonzero derivative, but that's quite restrictive.)

Reply 4

james22

16

Original post by krisshP

Let's say we have a straight line where an increase of x by 2 gives an increase in y by 5. Then

dy/dx=change in y/change in x
=5/2

dx/dy=change in x/change in y
=2/5

So, dy/dx * dx/dy =(5/2)*(2/5)=1, just a case of cancelling out.

For a general case,

dy/dx * dx/dy
=(change in y/change in x) * (change in x/change in y)
=1, just by cancelling.

dy/dx s not a fraction so you cannot do this. The proof is more complicated than that.

Reply 5

user2020user

16

Original post by james22

dy/dx s not a fraction so you cannot do this. The proof is more complicated than that.

Reply 6

MEPS1996

OP

4

can anybody tell me what the problem with my proof is?

Reply 7

Phichi

11

[br][br]\dfrac{1}{x}(x) = 1[br][br]

474aae8fe337f7e8f8c98838ab21056a49e4e299a604f10a55963ba54d03cd7d.jpg

(edited 10 years ago)

Reply 8

hassassin04

17

Original post by MEPS1996

can anybody tell me what the problem with my proof is?

How was the chain rule proved in the first place?

Reply 9

james22

16

Original post by MEPS1996

can anybody tell me what the problem with my proof is?

Because you are treating dy/dx as a fraction, which you cannot do.

Reply 10

Liamnut

13

Original post by james22

Because you are treating dy/dx as a fraction, which you cannot do.

Can't you say as y goes increase by n x increases by m.

dy/dx=n/m

dx/dy=m/n

So,

dy/dx*dx/dy=n/m*m/n=1

Maybe my teacher has simplified it all a bit too much or maybe I misunderstand her proof.

Reply 11

hassassin04

17

Original post by Liamnut

Can't you say as y goes increase by n x increases by m.

dy/dx=n/m

dx/dy=m/n

So,

dy/dx*dx/dy=n/m*m/n=1

Maybe my teacher has simplified it all a bit too much or maybe I misunderstand her proof.

It is not a proof but an intuitive way to look at it.

Reply 12

james22

16

Original post by Liamnut

Can't you say as y goes increase by n x increases by m.

dy/dx=n/m

dx/dy=m/n

So,

dy/dx*dx/dy=n/m*m/n=1

Maybe my teacher has simplified it all a bit too much or maybe I misunderstand her proof.

No, that is not a proof. You have basically assumed what you are trying to prove. If y=f(x), and g(x) is the inverse of f(x) (if f is not invertable then dx/dy doesn't really make sense anyway), then dx/dy=g'(y)=g'(f(x)). So what you need to prove is that g'(f(x))*f'(x)=1.

Reply 13

Smaug123

15

Original post by MEPS1996

can anybody tell me what the problem with my proof is?

According to your proof (as in my reply above),

\dfrac{\partial y}{\partial x}\dfrac{\partial z}{\partial y}\dfrac{\partial x}{\partial z}=1

, but it's actually

-1

.

Reply 14

FireGarden

14

Original post by Liamnut

Can't you say as y goes increase by n x increases by m.

dy/dx=n/m

dx/dy=m/n

So,

dy/dx*dx/dy=n/m*m/n=1

Maybe my teacher has simplified it all a bit too much or maybe I misunderstand her proof.

The problem you're having is a bit disguised when people say "you can't treat it as a fraction"; you have to treat it like:

\dfrac{dy(x_0)}{dx} = \displaystyle\lim_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}

which is the rigourous definition of the derivative. It is a limit, where now treating it as a fraction is no longer valid as the limit of the denominator is zero (otherwise, the 'algebra of limits' would have permitted treating it like a fraction - the issue is that if the denominator has a limit of zero, then 'as a fraction' it is undefined because we can't divide by zero).

Reply 15

MEPS1996

OP

4

Original post by hassassin04

How was the chain rule proved in the first place?

the limit of delta y/delta u * delta u/delta x was taken as delta x goes to 0

Reply 16

MEPS1996

OP

4

Original post by majmuh24

Because you have assumed what you are trying to prove.

Posted from TSR Mobile

No, i used the fact delta y/delta x * delta x/delta y=1 to prove dy/dx *dx/dy=1

Reply 17

cojocarutudor

Very interesting and useful tips.I read step by step this informations and help me very much.Thank you very much for this informations!

Reply 18

MEPS1996

OP

4

Original post by FireGarden

The problem you're having is a bit disguised when people say "you can't treat it as a fraction"; you have to treat it like:

\dfrac{dy(x_0)}{dx} = \displaystyle\lim_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}

which is the rigourous definition of the derivative. It is a limit, where now treating it as a fraction is no longer valid as the limit of the denominator is zero (otherwise, the 'algebra of limits' would have permitted treating it like a fraction - the issue is that if the denominator has a limit of zero, then 'as a fraction' it is undefined because we can't divide by zero).

im treating delta y/delta x as a fraction, not dy/dx, where delta represents a small but finite change

Reply 19

davros

16

Original post by MEPS1996

im treating delta y/delta x as a fraction, not dy/dx, where delta represents a small but finite change

Which is fine as long as you only deal with delta y and delta x.

But dy/dx and and dx/dy are not fractions, they are the result of processes - specifically, limiting processes.

Formally, we define:

\displaystyle \dfrac{dy}{dx} = \displaystyle \lim_{\delta x \to 0} \dfrac{\delta y}{\delta x}

which means "let delta-x go to 0 and consider the limit of the ratio of (delta y)/(delta x)" and:

\displaystyle \dfrac{dx}{dy} = \displaystyle \lim_{\delta y \to 0} \dfrac{\delta x}{\delta y}

which means "let delta-y go to 0 and consider the limit of the ratio of (delta x)/(delta y)"

Note that each of these involves a different limiting process - the former looks at what happens when we let delta-x tend to 0 as our independent variable; the latter looks at what happens when we let delta-y tend to 0 as our independent variable.

You're trying to argue that dy/dx and dx/dy "are just limits of fractions" without specifying how those limits are taken - the limit definition requires a statement of the thing you're allowing to vary!

That's why we need a bit of formal machinery (the Inverse Function Theorem) to tell us that what we'd like to happen is what really happens :smile:

(edited 10 years ago)

proof of dy/dx *dx/dy=1

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