# Confused at: du = f'(x) dx

1. Apr 23, 2007

### Sleek

Hello,

While solving integrals using substitution method, we often come across this,

if u = f(x), du = f'(x) dx

I would like to know if there exists a proof for the above equation. The problem is, I am totally dissatisfied by the explanation provided to me in the classes. During the derivatives classes, we were told that in case of (dy)/(dx), it actually means (d/dx)y, where (d/dx) stands as a notation, as opposed to dy being divided by dx.

The way the folks at my classes used substitution in the integral classes was,

let u = f(x)

diff. both sides w.r.t. x

du/dx = f'(x)

thus du = f'(x) dx [Multiplying by dx on both the sides]

But the last step's explanation seems to be too weird to me. How can a part of a notation be canceled off one side? Or is my understanding of the notation in the first place itself is wrong?

2. Apr 23, 2007

### ZioX

Technically, what you're saying is wrong. du is not f'(x)dx.

Where this comes from is when you prove the fundamental theorem of calculus you get a corollary that says this: g(b)=d, g(a)=c.

$$\int^d_cf(g(x))g'(x)dx=\int^{g(b)}_{g(a)}f(g(x))dx=\int^{g(b)}_{g(a)}f(u)du$$

where u=g(x). So from comparing the lhs and the rhs it's tempting to say that g'(x)dx=du.

Last edited: Apr 23, 2007
3. Apr 24, 2007

### Sleek

Thanks for the insight. The problem is, " du = f'(x) dx " is not something that I've assumed. Its currently being used in my class notes. Moreover, after searching various online sources, the reason for similar steps of substitution, "du = f'(x) dx ". Example of this would be: http://archives.math.utk.edu/visual.calculus/4/substitutions.3/. If you view the first sum, the reason for the resubstitution is given by the above equation I mentioned. The only reason I managed to find for that step is, "By DefN of Indefinite Integrals".

But I should thank you for the information you provided.

Regards,
Sleek.

4. Apr 24, 2007

### quasar987

You should regard the whole "du = f'(x) dx" buisiness as just a simplified notation for the "real" change of variable theorem evoked by our friend ZioX.

How it really works is that we look at the integrand of the integral and we ask... is there a function u(x) and a function f(x) such that the integrand is actually given by f(u(x))*u'(x) ?

So we start trying things; we say, ok let u(x)=... Then, du/dx=... If it works. That is to say, if there is indeed an f such that the integrand can be written f(u(x))*u'(x), then it means that we can switch the form of the integral to the one on the far right in ZioX's post. And what we notice is that u'(x)dx was actually replaced by just du.

So the net effect of all this was to change u'(x)dx by du. So why not immediately just abuse the weak damsel in distress that is the differential 'd' notation and write directly du=u'(x)dx instead of du/dx=... the first time we introduce the u substitution.

Last edited: Apr 24, 2007
5. Apr 24, 2007

### Gib Z

dy and dx are called differentials.
$\lim_{\Delta x \to 0} \Delta x = dx$ and similarly for dy.

Therefore dy/dx can be thought of as the ratio of 2 infinitesimal changes. That is why dy and dx mean nothing by themselves. Infinitesimals do not have numerical values. The ratio may. The only reason we can pretend dx and dy actually mean things by themselves, as if dy/dx is a fraction, is because by the definition of the derivative, it is the limit of a fraction.

My post would end well attaching quasars post to it :)

6. Apr 24, 2007

### HallsofIvy

Staff Emeritus
You are very correct that "dy/dx" is NOT a fraction. When we first define the derivative, we do NOT define "dy" and "dx" separately. However, it is true that dy/dx is the limit of a fraction and so can be "treated" like a fraction. That is, the chain rule, dy/dz= (dy/dx)(dx/dz) cannot be proved by saying "just cancel the dx's"! But you can prove it by going back before the limit and canceling the corresponding things there. That is, again, although dy/dx is not a fraction, we can always treat it as if it were.

That's why, in most calculus text books, after we have defined the derivative as a "limit of fractions", we define the differentials, dy and dx, separately: by declaring that "dx" is a symbol and then dy by dy= f'(x)dx. Thus, there is no "assumption" or "proof" that dy= f'(x)dx- it is a definition- of dy.

7. Apr 24, 2007

### ObsessiveMathsFreak

du=f'(x)dx is, despite the fact that it works fairly well, strictly speaking not correct. OK, we all use it and it gets results, but it's really just a shorthand for a more complete explanation.

Without resorting to integral formulae, you can do the following to justify the construction.

You have u=f(x). First, let x=x(t), in other words, x is now a function of another variable t. With a little thought, you can now see that u is also a function of this new variable.

u=f(x(t))

Now differentiate u with respect to t

$$\frac{du}{dt} = \frac{df}{dx}\frac{dx}{dt}$$

$$\frac{du}{dt} = f'(x(t))\frac{dx}{dt}$$

Now many would say that the dt's cancel(they don't) and you get du=f'(x(t))dx or du=f'(x)dx if you make x an independant variable again.

But the dt's do not cancel. People have invented entirely new mathematical formalisms in order to be able to cancel or just get rid of those dt's in this way, but since we're not dealing with such formalisms, we have to leave them in. (They're inseperable from the upper d anyway.)

In conclusion you're supposed to use
$$\frac{du}{dt} = f'(x(t))\frac{dx}{dt}$$
But everyone uses
$$du = f'(x)dx$$
for short. Unfortunately the shorthand has led to a lot of confusion, which is why the original question was asked.

8. Apr 24, 2007

### AiRAVATA

You shall all listen to HallsofIvy. $dy$ is a well defined mathematical object called differential, and it goes beyond the change of variables method or integrals. There is a tight theory regarding this objects and is more than correct to write $dy=f'(x)dx$.

Last edited: Apr 24, 2007
9. Apr 24, 2007

### HallsofIvy

Staff Emeritus
But a "symbolic" object rather than a "real" object in much the same way that "$\nabla$", while not a "real" vector is a very useful object!

The real strength of "differentials" comes in differential geometry where they are very strictly defined.

10. Apr 24, 2007

### Crosson

We had a professor come from Bulgaria, she is very friendly and teaching-oriented, but when she saw how we teach "u-substitution" for change of variables under the integral sign she was shocked.

11. Apr 24, 2007

### Sleek

Thanks for all of your replies. I can now better understand what happens during these substitutions. It has cleared most of my doubts.

Thanks again!

Regards,
Sleek.

12. Apr 25, 2007

### jostpuur

I have traumatic experience on this stuff. If I was a dictator, infinitesimal differentials would be forbidden by the law. Those infinitesimals are often defended for their intuitive value, but my lecturer at least only nearly attempted to confuse us as much as he could, when explained about these differentials. Potential for abuse is huge.

Anyway, just in case somebody who's still learning the basics happened to see those equations, I must correct them. $$\lim_{\Delta x\to 0}\Delta x = dx$$, is not correct, but $$\lim_{\Delta x\to 0}\Delta x = 0$$ is.

13. Apr 25, 2007

### Gib Z

Ahh...define dx in terms of delta x for me then? I know the rigourous way to think of them, but this guys a newbie, give him some intuition.

14. Apr 25, 2007

### HallsofIvy

Staff Emeritus
You don't. dx is not define "in terms of delta x" (although some texts use "dx" when they should use delta x). If you think of "dx" as an infinitesmal then delta x is approximately dx.

"dx" is a symbol representing an infinitesmal. dy is then defined as f'(x)dx.

Notice the difference between 'an infintesmal' and 'representing an infintesmal'! If you are going to think of dx as being an infinitesmal, then you had better give a rigorous definition of "infinitesmal". As long as you are thinking of dx as representing an infinitesmal, you don't have to!