# Total derivative?

1. Jun 28, 2011

### Lindsayyyy

1. The problem statement, all variables and given/known data

Is the following equation a total derivative?

$$dz = 2ln(y)dx+{\frac x y}dy$$

2. Relevant equations
-

3. The attempt at a solution

I would say no. I tried it with the symmetry of the second derivatives.

$$2ln(y)$$ is $${\frac {\partial z} {\partial x}}$$

when I derivate it now and say y is my variable I get:

$${\frac {\partial z} {\partial y}}=\frac 2 y$$

same for the other expression

$${\frac {\partial z} {\partial y}}=\frac x y$$

$${\frac {\partial z} {\partial x}}=\frac 1 y$$

so they are different => no totale derivative. Am I right or where did I do mistakes?

Thanks for the help

2. Jun 28, 2011

### tiny-tim

Hi Lindsayyyy!

That's completelyyyy right.

(just to check, I like to differentiate the obvious possibility, xlny, and that gives lnydx + xdy/y, which doesn't match)

3. Jun 28, 2011

### Lindsayyyy

Thank you very much

4. Jun 28, 2011

### HallsofIvy

Staff Emeritus
Just to put my oar in, if that were a total derivative, there would exist a function, F(x,y) such that
$$\frac{dF}= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy$$
$$\frac{dF}= 2 ln(y)dx+ \frac{x}{y}dy$$
..
So we must have $\partial F/\partial x= 2 ln(y)$ so $F= 2x ln(y)+ \phi(y)$ where $\phi(y)$ can be any differentiable function of y only.

From that $\partial F/\partial y= 2x/y+ \phi'(y)$ and that must be equal to $x/y$: $2x/y+ \phi'(y)= x/y$ so $\phi'(y)= -x/y$ which is impossible since $\phi'(y)$ is not a function of y.

Note the "mixed second derivatives". If $\partial F/\partial y= x/y+ \phi'(y)$
then $\partial^2 F/\partial x\partial y= 1/y$ and if $\partial F/\partial x= 2ln(y)$ then $\partial^2 F/\partial y\partial x= 2/y$. Those mixed second derivatives should be the same and that was what you were checking.

5. Jun 28, 2011

### Lindsayyyy

your solution is way more elegant than mine

I have another problem:

$$x^{\alpha}y^{\beta}(2ln(y)dx+{\frac x y}dy)$$ is given. I shall find alpha and beta that the given expression is a total derivative.

My approach:

I found $${ \frac {\partial^{2}z} {\partial x \partial y}}=2\beta x^{\alpha}y^{\beta -1}ln(y)+{\frac 1 y}2x^{\alpha}y^{\beta}$$

and
$${ \frac {\partial^{2}z} {\partial y \partial x}}=\alpha x^{\alpha -1}y^{\beta}{\frac x y}+{\frac {x^{\alpha}y^{\beta}} y}$$

I though I could equate them and solve the equation to get an expression for alpha and beta. But my problem is I can easily cancel some terms with potential rules like y^a-1=y^a/y etc.

And I get to the euqation:

$$2\beta lny+2 = \alpha+1$$

is my approach wrong, where are my mistakes? If I was right thus far, how should I go on?

Thanks for your help

6. Jun 28, 2011

### SammyS

Staff Emeritus
There is no y on the right hand side, so what must β be ?

7. Jun 28, 2011

### Lindsayyyy

fail :yuck:

$$\beta =0 \alpha =1$$

thanks, I'm blind, I know it's a bit too much and insolent to ask but you haven't calculated it by yourself in order to test if I was right until the last step, have you?

Thanks for your help

edit:

nevermind, I'm tired^^
I haven't thought about just to put it in and derive it haha. It works. thanks

Last edited: Jun 28, 2011
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