# Product rule - differentiation

#### amilton542

Joined Nov 13, 2010
497
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'

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#### Zazoo

Joined Jul 27, 2011
114
Yes, that works.

#### Georacer

Joined Nov 25, 2009
5,181
Are the rest of the functions related to x?

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

#### amilton542

Joined Nov 13, 2010
497
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?

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#### amilton542

Joined Nov 13, 2010
497
Ohhhh rats! I've made an error in the alphabet! How embarrassing. I've used y as notation in the functions, sorry folks!

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#### Zazoo

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

#### amilton542

Joined Nov 13, 2010
497
As Georacer pointed out, the assumption here is that u, v, w and z are all functions of x themselves (and not separate variables.)
When did I say they were seperate variables? An increment in x will cause incremental changes in all functions.

#### Zazoo

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