# Product rule - differentiation

Discussion in 'Math' started by amilton542, Feb 2, 2012.

1. ### amilton542 Thread Starter Active Member

Nov 13, 2010
494
64
Let's say you've got a product of n different functions, as many as you want it doesn't matter (but not two), I'll choose 5 for simplicity. If I wanted to calculate dy/dx, could I use this method?

(uvwxz)'= u'vwxz + uv'wxz + uvw'xz + uvwx'z + uvwxz'

Last edited: Feb 2, 2012
2. ### Zazoo Member

Jul 27, 2011
114
43
Yes, that works.

amilton542 likes this.
3. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
Are the rest of the functions related to x?

What you wrote is the derivative of the product. Do you want dy/dx instead?

4. ### amilton542 Thread Starter Active Member

Nov 13, 2010
494
64
Yes I wanted the derivative of the product.

I introduced all the new y values and subtracted the old ones to give me the difference in y, then I took the limit when Δx tends to zero, then put it in that notation for a clear visual inspecton.

Please excuse my terminology, I thought dy/dx was the total rate of instantaneous change e.g. the derivative which can alse be written as (uvwxz)'. Is that correct?

Last edited: Feb 2, 2012
5. ### amilton542 Thread Starter Active Member

Nov 13, 2010
494
64
Ohhhh rats! I've made an error in the alphabet! How embarrassing. I've used y as notation in the functions, sorry folks!

Last edited: Feb 2, 2012
6. ### Zazoo Member

Jul 27, 2011
114
43
As Georacer pointed out, the assumption here is that u, v, w and z are all functions of x themselves (and not separate variables.)

7. ### amilton542 Thread Starter Active Member

Nov 13, 2010
494
64
When did I say they were seperate variables? An increment in x will cause incremental changes in all functions.

8. ### Zazoo Member

Jul 27, 2011
114
43
You didn't, I was just pointing out that my answer was based on that assumption.