Correct Algebraic Manipulation of Partial Derivative Operations

In summary, it appears as though it is possible to rearrange the algebraic structure of a partial derivative problem in order to make it simpler. However, this is not always the case and it may not produce a simpler problem.
  • #1
Liquid7800
76
0

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
 
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  • #2
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.
 
  • #3
Liquid7800 said:

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. fxy does not mean (fx)*(fy). It means [tex]\frac{\partial}{\partial y}\frac{\partial f}{\partial x}[/tex]

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

[tex]f_{xy} = \frac{\partial}{\partial y}f_x = 2[/tex]

On the other hand, fx * fy = (2x + 2y) * 2x = 4x2 + 4xy.
Liquid7800 said:
...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
 
  • #4
Thanks Mark,

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

I know that:

[tex]
f_{xy} = \frac{\partial}{\partial y}f_x = 2
[/tex]

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.
 
  • #5
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 ...?
 

1. What is partial differentiation?

Partial differentiation is a mathematical operation that involves finding the derivative of a function with respect to one of its independent variables while holding all other variables constant.

2. Why is it important to correctly manipulate partial derivatives?

Correct manipulation of partial derivatives is crucial in many fields of science and engineering, as it allows for the precise calculation of rates of change and optimization of functions with multiple variables.

3. How do you correctly manipulate partial derivatives?

To correctly manipulate partial derivatives, you must first identify the variables in the function and their respective exponents. Then, use the power rule to find the derivative of each term, and multiply the resulting derivatives by the corresponding original variables.

4. What are some common mistakes when manipulating partial derivatives?

Some common mistakes when manipulating partial derivatives include forgetting to apply the chain rule when dealing with nested functions, not correctly identifying the variables and their exponents, and incorrectly using the product or quotient rule.

5. What are some applications of partial derivatives in science?

Partial derivatives have many applications in science, such as in physics for calculating rates of change in thermodynamics and fluid dynamics, in economics for optimizing production and cost functions, and in biology for modeling population growth and reaction rates.

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