Is Equation a Total Derivative?

In summary, the conversation discusses whether or not a given equation is a total derivative and how to find the values for the variables that would make it a total derivative. The conclusion is that the equation is not a total derivative and the values of α and β are 1 and 0, respectively.
  • #1
Lindsayyyy
219
0

Homework Statement



Is the following equation a total derivative?

[tex] dz = 2ln(y)dx+{\frac x y}dy[/tex]

Homework Equations


-


The Attempt at a Solution



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

[tex] 2ln(y) [/tex] is [tex] {\frac {\partial z} {\partial x}}[/tex]

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

[tex] {\frac {\partial z} {\partial y}}=\frac 2 y[/tex]

same for the other expression

[tex] {\frac {\partial z} {\partial y}}=\frac x y[/tex]

[tex] {\frac {\partial z} {\partial x}}=\frac 1 y[/tex]


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

Thanks for the help
 
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  • #2
Hi Lindsayyyy! :wink:

That's completelyyyy right. :smile:

(just to check, I like to differentiate the obvious possibility, xlny, and that gives lnydx + xdy/y, which doesn't match)
 
  • #3
Thank you very much:smile:
 
  • #4
Just to put my oar in, if that were a total derivative, there would exist a function, F(x,y) such that
[tex]\frac{dF}= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy[/tex]
[tex]\frac{dF}= 2 ln(y)dx+ \frac{x}{y}dy[/tex]
..
So we must have [itex]\partial F/\partial x= 2 ln(y)[/itex] so [itex]F= 2x ln(y)+ \phi(y)[/itex] where [itex]\phi(y)[/itex] can be any differentiable function of y only.

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

Note the "mixed second derivatives". If [itex]\partial F/\partial y= x/y+ \phi'(y)[/itex]
then [itex]\partial^2 F/\partial x\partial y= 1/y[/itex] and if [itex]\partial F/\partial x= 2ln(y)[/itex] then [itex]\partial^2 F/\partial y\partial x= 2/y[/itex]. Those mixed second derivatives should be the same and that was what you were checking.
 
  • #5
your solution is way more elegant than mine :smile:

I have another problem:

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

My approach:

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

and
[tex]{ \frac {\partial^{2}z} {\partial y \partial x}}=\alpha x^{\alpha -1}y^{\beta}{\frac x y}+{\frac {x^{\alpha}y^{\beta}} y} [/tex]

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:

[tex] 2\beta lny+2 = \alpha+1[/tex]

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

Thanks for your help
 
  • #6
Lindsayyyy said:
...
And I get to the euqation:

[tex] 2\beta lny+2 = \alpha+1[/tex]

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

Thanks for your help
There is no y on the right hand side, so what must β be ?
 
  • #7
fail :yuck:

[tex] \beta =0 \alpha =1 [/tex]

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:

1. What is a total derivative?

A total derivative is a mathematical concept that represents the instantaneous rate of change of a multivariable function with respect to its independent variables. It takes into account all partial derivatives of the function and is often denoted by the symbol d.

2. How do you determine if an equation is a total derivative?

To determine if an equation is a total derivative, you can use the total derivative definition which states that if the equation satisfies the condition of having all partial derivatives exist and be continuous, then it is a total derivative. Additionally, you can use the total derivative test to check if the equation satisfies the conditions for being a total derivative.

3. What is the difference between a total derivative and a partial derivative?

A partial derivative only considers the change in one variable while holding all other variables constant, whereas a total derivative considers the change in all variables simultaneously. In other words, a total derivative takes into account the effect of all independent variables on the function, while a partial derivative only looks at the effect of one variable at a time.

4. Can an equation be both a total derivative and a partial derivative?

Yes, an equation can be both a total derivative and a partial derivative. This can happen when the equation satisfies the conditions for both types of derivatives. In this case, the equation would be considered a total derivative with respect to all variables and a partial derivative with respect to one variable.

5. What are the applications of total derivatives?

Total derivatives have many applications in fields such as physics, economics, and engineering. They are used to calculate instantaneous rates of change, optimize functions, and approximate solutions to complex problems. Total derivatives also play a crucial role in multivariable calculus and are essential for understanding higher-level mathematical concepts.

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