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Topic: total differentials  (Read 12028 times)

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Offline zeshkani

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total differentials
« 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

Offline Yggdrasil

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Re: total differentials
« Reply #1 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.
« Last Edit: August 30, 2007, 03:36:17 PM by Borek »

Offline zeshkani

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Re: total differentials
« Reply #2 on: August 30, 2007, 11:27:30 PM »
would this be the answer:

dz=2x-2y+2dx + (4y-2x-4 dy)
« Last Edit: August 30, 2007, 11:47:39 PM by zeshkani »

Offline Yggdrasil

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Re: total differentials
« Reply #3 on: August 31, 2007, 02:18:19 AM »
Do you mean:

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

Offline zeshkani

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Re: total differentials
« Reply #4 on: August 31, 2007, 10:32:25 AM »
yeah thats what i mean i just forgot to put () around the dx part

Offline FeLiXe

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Re: total differentials
« Reply #5 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
Math and alcohol don't mix, so... please, don't drink and derive!

Offline zeshkani

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Re: total differentials
« Reply #6 on: August 31, 2007, 09:20:01 PM »
thx for all the help, i finally got it  ;D

Offline Yggdrasil

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Re: total differentials
« Reply #7 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).

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