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Specialty Chemistry Forums => Other Sciences Question Forum => Topic started by: zeshkani on August 30, 2007, 01:49:21 PM

Title: total differentials
Post by: zeshkani on August 30, 2007, 01:49:21 PM

can somebody explain on how to do a total differentiation on this, i have the formula but iam just lost

z=x^2+2y^2-2xy+2x-4y-8

Title: Re: total differentials
Post by: Yggdrasil on August 30, 2007, 02:36:48 PM

You can look at this term by term to make things simpler.  I'll do the first couple of terms to start things off:

x²
The differential of this term is fairly easy since it involves only one variable.  I like to think of a differential this way.  You know the derivative (dz/dx) of z = x²:

(dz/dx) = 2x

Well, if you treat the derivative like a fraction, and multiply through by dx, you get:

dz = 2x dx

which is the differential of x².


-2xy
This term is a more difficult term.  Here you have to make use of the product rule:

d(uv) = u dv + v du

So in this case,

d(-2xy) = -2 d(xy) = -2(xdy + ydx) = -2x dy - 2y dx

You should be able to do the rest now.

Title: Re: total differentials
Post by: zeshkani on August 30, 2007, 11:27:30 PM

would this be the answer:

dz=2x-2y+2dx + (4y-2x-4 dy)

Title: Re: total differentials
Post by: Yggdrasil on August 31, 2007, 02:18:19 AM

Do you mean:

dz = (2x-2y+2) dx + (4y-2x-4) dy

Title: Re: total differentials
Post by: zeshkani on August 31, 2007, 10:32:25 AM

yeah thats what i mean i just forgot to put () around the dx part

Title: Re: total differentials
Post by: FeLiXe on August 31, 2007, 02:23:21 PM

if you don't want to think like Yggdrasil you could form the partial derivatives:
dz/dx = 2x - 2y + 2
dz/dy = 4y - 2x - 4

then the differential is defined as:
dz = (2x - 2y + 2)dx + (4y - 2x - 4)dy

Title: Re: total differentials
Post by: zeshkani on August 31, 2007, 09:20:01 PM

thx for all the help, i finally got it  ;D

Title: Re: total differentials
Post by: Yggdrasil on September 01, 2007, 12:07:37 AM

FeLiXe's method is probably the easier way to think of things.  Conceptually, though it is important to know the product rule for differentials (i.e. d(uv) = u dv + v du).