# Total Differential Problem

• Jun 1st 2012, 12:33 PM
econolondon
Total Differential Problem
Hi

I am trying to get the total differential for

y = x1x2/2x1 + x2

I have been given the answer

dy = [x2/2x1 + x2]dx1 - [2x1x2/(2x1+x2)^2]dx1 + [x1/2x1 + x2]dx2 - [x1x2/(2x1+x2)^2]dx2

Is this right?

It is my understanding that the general formula is F1dx1 + F2dx2, for y=F(x1,x2)

I don't understand why, in the second and fourth terms of the answer, the WHOLE of the denominator is being squared, rather than just the x1 or x2 that the partial derivative refers to......
• Jun 1st 2012, 09:54 PM
earboth
Re: Total Differential Problem
Quote:

Originally Posted by econolondon
Hi

I am trying to get the total differential for

y = x1x2/2x1 + x2

I have been given the answer

dy = [x2/2x1 + x2]dx1 - [2x1x2/(2x1+x2)^2]dx1 + [x1/2x1 + x2]dx2 - [x1x2/(2x1+x2)^2]dx2

Is this right?

It is my understanding that the general formula is F1dx1 + F2dx2, for y=F(x1,x2)

I don't understand why, in the second and fourth terms of the answer, the WHOLE of the denominator is being squared, rather than just the x1 or x2 that the partial derivative refers to......

1. I assume that your function reads: $\displaystyle \displaystyle{y = \frac{x_1 \cdot x_2}{2x_1+x_2}}$

2. You have to use the quotient rule: $\displaystyle f = \frac uv~\implies~ f'=\frac{v \cdot u' - u \cdot v'}{v^2}$

3. I'll show you the $\displaystyle F_1dx_1$ part:

The variable is $\displaystyle x_1$. $\displaystyle x_2$ is considered to be constant.

$\displaystyle F_1dx_1 = \frac{(2x_1+x_2) \cdot x_2 - x_1 \cdot x_2 \cdot 2}{(2x_1+x_2)^2}$

4. The $\displaystyle F_2dx_2$ part has to be done in excatly the same way.
• Jun 2nd 2012, 03:07 AM
econolondon
Re: Total Differential Problem