Total Differential Problem

Hi

I am trying to get the total differential for

y = x_{1}x_{2}/2x_{1} + x_{2}

I have been given the answer

dy = [x_{2}/2x_{1} + x_{2}]dx_{1} - [2x_{1}x_{2}/(2x_{1}+x_{2})^2]dx_{1} + [x_{1}/2x_{1} + x_{2}]dx_{2 }- [x_{1}x_{2}/(2x_{1}+x_{2})^2]dx_{2 }

Is this right?

It is my understanding that the general formula is F_{1}dx_{1} + F_{2}dx_{2}, for y=F(x_{1},x_{2})

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 x_{1 }or x_{2} that the partial derivative refers to......

Re: Total Differential Problem

Quote:

Originally Posted by

**econolondon** Hi

I am trying to get the total differential for

y = x_{1}x_{2}/2x_{1} + x_{2}

I have been given the answer

dy = [x_{2}/2x_{1} + x_{2}]dx_{1} - [2x_{1}x_{2}/(2x_{1}+x_{2})^2]dx_{1} + [x_{1}/2x_{1} + x_{2}]dx_{2 }- [x_{1}x_{2}/(2x_{1}+x_{2})^2]dx_{2 }

Is this right?

It is my understanding that the general formula is F_{1}dx_{1} + F_{2}dx_{2}, for y=F(x_{1},x_{2})

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 x_{1 }or x_{2} 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.

Re: Total Differential Problem

Thanks, that's very helpful.