Question

Chain rule question

Answer 1

\[\frac{ dy }{ dt }=\frac{dy}{ dx }\frac{ dx }{dt}\]

Why can't we cross out the dx's like a fraction?

Surely if Jenny runs twice as fast as Carl and Carl runs three times as fast as Annie, then Jenny runs six times as fast as Annie?

Answer 2

where is the question

Answer 3

"Why can't we cross out the dx's like a fraction?"

Answer 4

In the middle.

Answer 5

Because rate of change isn't a fraction, it can just sometimes be treated like one.

Answer 6

When can't it be treated as a fraction? Seems like rate of change is always Something per Something Else.

Answer 7

I just don't understand why there is ever any hesitation here on treating dy/dx as a fraction. I don't get it. I've done calculus for several years now and I still just don't get it. Using it as a fraction has always practically worked for me and in fact is part of my intuition for why it works... I just don't get why people seem to say it's not...

Answer 8

http://mathoverflow.net/questions/73492/how-misleading-is-it-to-regard-fracdydx-as-a-fraction

Answer 9

Hey dude, we DO treat them as fractions here! That is the point of this rule.

Answer 10

dy/dt can be written as dy/dx * dx/dt since the dx's get cancelled and we are left with dy/dt.

Answer 11

So if we know dy/dx and dx/dt, we can multiply them and get dy/dt.

Answer 12

dy/dx is a fraction. And it is to be treated as one.

Answer 13

This isn't really what I'm looking for @ParthKohli since you're just agreeing with me and not teaching me anything.

See, I want to understand why someone would believe that they can't be treated as fractions.

@agent0smith I just finished reading what you linked me to, and I didn't see anything that made me question it. In fact, one thing I found interesting was:

Suppose you say x=tan(theta)

Then\[\frac{ dx }{ d \theta }=\sec^2 \theta = 1+\tan^2 \theta=1+x^2\]

Then if you treat it like a fraction, \[\frac{ d \theta }{ dx }=\frac{ 1 }{ 1+x^2 }\]

Which if you integrate that with respect to x you get the arctangent function where we started from.

--

I need to see a concrete example where treating differentials as fractions fails.

Answer 14

You could try this one: http://math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio

Answer 15

I kind of am tired of being referred elsewhere, I suppose I just want to see a result where it is done and fails, instead of treading through more of the same kind of stuff. Just skimming through what you've sent me seems to be even more in support of the affirmative, that you can treat them as fractions... Can you direct me to a specific part in there that I should start at?

Answer 16

Sorry to be difficult, I just specifically want to see this fail, cause I can't believe it until I see it.

Answer 17

I think the reason you're not seeing it fail is because it works as a fraction at most of the lower levels of calculus. I can't give a specific example, since I'm not familiar enough with the upper levels of calculus anymore.

Answer 18

can we introduce partials ?

Answer 19

consider, \(y = xz\)

compute \(\frac{\partial y}{dx}\) and \(\frac{\partial y}{dz}\), and try to take their ratio treating these "weird objects" as fractions.
next, compute \(\frac{\partial z}{dx}\) implicitly from the funciton. you may see why these things should not be treated as fractions sometimes.

Answer 20

all symbols are \(\partial\) above...

Answer 21

@ganeshie8 you should use \Large at the front if you want to make it easier to read \Large eg.
\Large \frac{\partial y}{ d x}
[\Large \frac{\partial y}{ d x} \]

Answer 22

oops \[\Large \frac{\partial y}{ d x} \]

Answer 23

yah ikr :)

http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-11-chain-rule/

watch 2 minutes of this video between :-
6:00 - 8:00

Answer 24

it talks about differentials. not derivatives.
still i thought its good to go thru that once hmm

Answer 25

I am not seeing it, sorry. I think manipulating partial derivatives like you are is using them is out of context, since they are really already related by something like this, you can't just ignore it:

\[dz=\frac{ \partial z }{ \partial x } dx + \frac{ \partial z }{ \partial y }dy\]

But I was never really concerned with partial derivatives in my question to begin with.

Answer 26

then if u limit urself to chain rule, i see no problem either

Answer 27

What I'm saying is, I know that this isn't true always: \[\frac{ dy }{ dx }=\frac{ \partial y }{ \partial x }\]

But I don't think that's why people say you can't cross out infinitesimals like fractions, just because people might confuse "d" with a squiggly "d" for the same thing.

Answer 28

*then if u limit urself to chain rule in single variable calculus, i see no problem either

Answer 29

**face hits desk**

I conclude that Real Analysis is a waste of time and complete garbage.

Answer 30

lol, I like below justification from SE :-

So Calculus was essentially rewritten from the ground up in the following 200 years to avoid these problems, and you are seeing the results of that rewriting (that's where limits came from, for instance). Because of that rewriting, the derivative is no longer a quotient, now it's a limit:

\(\lim \limits_{h→0} \frac{f(x+h)−f(x)} {h} \)

And because we cannot express this limit-of-a-quotient as a-quotient-of-the-limits (both numerator and denominator go to zero), then the derivative is not a quotient.

http://math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio

Answer 31

I don't believe it. Just because you can't write it as a quotient of limits doesn't mean it's not a quotient, sorry. All this says is that you are scaling both parts of the quotient together since they're related... Obviously... Painfully...

Answer 32

yes, interesting... guess i see what u mean.
il have to think bit more thoroughly; idk how being able to write the limit as ratio-of-limits, implies the quantity to be a quotient and vice-versa :o

Answer 33

Thinking upon this a little more I feel like this is completely a human fault anyways. The quotient of the limits of something like f(x)=x^2 will be just 0/0 which is undefined. It's almost like every time you have something undefined you can say the same thing with a limit to define it.

Why go through the trouble and just say that all these things are defined with a "partner function".

It's like when someone asks you how tall you are and someone says, 3. Three feet? Three yards? Three meters? It just feels like this so much and has always bothered me.

Answer 34

That and people trying to "define" things in Analysis as though they can somehow claim to "understand" infinity and infinitesimally small. Pffft, the whole thing is garbage based on fake mind games that is only capable of recreating things we already intuitively understand that were in actuality discovered as a part of reality and never defined in the first place. It really just feels like 100% garbage.

Answer 35

Well it was fun. I'm ready for next semester to start already so I can learn some more calculus, I'm taking Diff Eq 2 and Vector Cal. =D