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Why do they work: chain rule, product rule, integration by parts?
By Olof Sisask (P3033) on Wednesday, October 25, 2000 - 12:51 pm:

Hi,

I was wondering what the reasoning is behind the Chain rule and the Product rule - why they work rather than how. Also, what is integration by parts? I would really appreciate some help on this, as I can't get ahold of any good books on the subject.

Thank you,
Olof.

By Susan Langley (Sml30) on Wednesday, October 25, 2000 - 01:33 pm:

Ok, the chain rule. You start by differentiating y in terms of x: yes, that is dy/dx. Now, consider u, and u is a function of x, say u(x). We can differentiate u in terms of x, thats du/dx and y in terms of u, thats dy/du.

OK, so what does that achieve? So, what is dy/dx? It's the rate of change of y to x in terms of x. So dy/du is the rate of change of y to u in terms of u, which, if you write u in terms of x becomes the rate of change of y to u in terms of x. And du/dx is the rate of change of u to x in terms of x. So isn't dy/du × du/dx the rate of change of y to x in terms of x, or dy/dx.

Ok that's the chain rule. I think I have a nice explanation on the product rule and possibly integration by parts in my room. So, I'll send you those when I get a chance to get back to the computer rooms...

By Olof Sisask (P3033) on Wednesday, October 25, 2000 - 01:47 pm:

Thanks Susan!

By Susan Langley (Sml30) on Wednesday, October 25, 2000 - 05:49 pm:


...Ok, here's product rule:

Consider this diagram:

diagram

y
=
uv
 dy
dx
=

lim
dx®0 
  (u+du)(v+dv) - uv
dx
=

lim
dx®0 
  udv + vdu - dudv
dx
=

lim
dx®0 
 æ
è		 u  dv
dx
 + v  du
dx
 + du  dv
dx
 ö
ø
=
u  dv
dx
 + v  du
dx

And, I need to go again - integration by parts to follow...

By Olof Sisask (P3033) on Wednesday, October 25, 2000 - 06:39 pm:

Thanks again Susan. Couple of questions:
Does lim(dx®0) mean the limit of <whatever follows> as dx tends to 0?
Also, what happens to the + du(dv/dx) part?

Thanks in advance,
Olof.

By Brad Rodgers (P1930) on Wednesday, October 25, 2000 - 08:33 pm:

As du tends to 0, because dv/dx is a finite value, call it x, 0×x=0. So it tends to 0 and disappears. You are right about limits.

Brad

By Olof Sisask (P3033) on Wednesday, October 25, 2000 - 09:33 pm:

Ah I see, cheers Brad.

By Susan Langley (Sml30) on Thursday, October 26, 2000 - 09:50 am:

OK, integration by parts, which uses the product rule:

uv
=
ó
õ		 x

x0 
  d(uv)
dx
 dx
=
ó
õ		 x

x0 
  du
dx
 v dx + ó
õ		 x

x0 
 u  dv
dx
 dx
so ó
õ		 x

x0 
 u  dv
dx
 dx
=
uv - ó
õ		 x

x0 
  du
dx
 v dx

Does that help?
Susan
By Olof Sisask (P3033) on Tuesday, November 7, 2000 - 04:14 pm:

Cheers guys!

Regards,
Olof