# Math Help - Total Differential Problem

1. ## 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......

2. ## Re: Total Differential Problem

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: $\displaystyle{y = \frac{x_1 \cdot x_2}{2x_1+x_2}}$

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

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

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

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

4. The $F_2dx_2$ part has to be done in excatly the same way.