Correct Algebraic Manipulation of Partial Derivative Operations

[Liquid7800]

Homework Statement

Hello, recently in my calculus III class we went over some problems of the following form:
If 'for some given equation' show:
x(fx) +y(fxy) +fx= 0

For some examples I was playing around with the operations BEFORE I dived in and started solving for fx and fy and got the correct answers for some problems----so I thought Now is it possible to 're-arrange' the algebraic structure as such for the partial derivative operations?

x(fx) + y(fx(f+fy))=0 ; where 'f' is the undifferentiated function?

...in the same manner can you 'divide' out the partial derivative operations too? Therefore getting

Consider from the same equation above,
x(fx) = -(y)(fx)(f + fy)

We divide out fx
as we solve for 'x' getting:

x = -(y)(f + fy)

We then place 'x' back into the original equation and show that
-(y)(f + fy) = -(y)(f + fy)

Therefore to 'complete' the proof we just solve for fy and add it to f; where 'f' is the function; multiplied by 'y' and show that they do indeed equal 0. Thus allowing only to solve for fy to prove the problem?

Now re-arranging this problem algebraically may not make things any simpiler but sometimes it may AND it at least helps me see the big picture.

Is it possible to perform these kind of algebraic manipulations on these sort of 'proof' problems?

Thanks for your help as always

 

[Liquid7800]

Ive never been one to BUMP a thread but, I really would like to know if operations on problems like these are 'legal'----I mean do these type of problems follow the elementary algebraic operations? In addition, let me know if my question does not make sense.
Thanks again.

 

[Mark44]

Homework Statement

Hello, recently in my calculus III class we went over some problems of the following form:
If 'for some given equation' show:
x(fx) +y(fxy) +fx= 0

For some examples I was playing around with the operations BEFORE I dived in and started solving for fx and fy and got the correct answers for some problems----so I thought Now is it possible to 're-arrange' the algebraic structure as such for the partial derivative operations?

x(fx) + y(fx(f+fy))=0 ; where 'f' is the undifferentiated function?

This is not correct. You are confusing the differentiation operators with the functions produced by differentiation. f_xy does not mean (f_x)*(f_y). It means \frac{\partial}{\partial y}\frac{\partial f}{\partial x}

For example, suppose f(x, y) = x² + 2xy. Then f_x = 2x + 2y, and f_y = 2x.

f_{xy} = \frac{\partial}{\partial y}f_x = 2

On the other hand, f_x * f_y = (2x + 2y) * 2x = 4x² + 4xy.

...in the same manner can you 'divide' out the partial derivative operations too? Therefore getting

Consider from the same equation above,
x(fx) = -(y)(fx)(f + fy)

We divide out fx
as we solve for 'x' getting:

x = -(y)(f + fy)

We then place 'x' back into the original equation and show that
-(y)(f + fy) = -(y)(f + fy)

Therefore to 'complete' the proof we just solve for fy and add it to f; where 'f' is the function; multiplied by 'y' and show that they do indeed equal 0. Thus allowing only to solve for fy to prove the problem?

Now re-arranging this problem algebraically may not make things any simpiler but sometimes it may AND it at least helps me see the big picture.

Is it possible to perform these kind of algebraic manipulations on these sort of 'proof' problems?

Thanks for your help as always

 

[Liquid7800]

Thanks Mark,

However perhaps I wasnt as clear as I needed to be...

I know that:

<br /> f_{xy} = \frac{\partial}{\partial y}f_x = 2<br />

and

On the other hand, fx * fy = (2x + 2y) * 2x = 4x2 + 4xy.

...what I meant was re-arranging the ORDER of partial differentiation with respect to the 'respective' variable:



Not multiplying them together...

for example, I got a problem:

Consider,
∂²z/∂x² + 2(∂²z/∂x∂y) + ∂²z/∂y²

now is :
∂²z/∂x² + 2(∂²z/∂x∂y) + ∂²z/∂y² ≡ (∂z/∂x + ∂z/∂y)²

This what I meant by simplifying the total number of operations when considering the problem...in this case possibly using an identity-ish simplification.

 

[Liquid7800]

so is this like a similar identity to

(a+b)^2 == a^2 +2ab + b^2 ?

and in the same way could it be extended to (a+b)^n ...?