Do we treat x and y as independent when differentiating f with respect to y?

In summary: Then you need to use Gateaux derivatives :f(x(t)+hn(t),y(t))=x(t)^2+y(t)^2+2x(t)hn(t)+h^2n(t)^2and thus :D_{x(t),n(t)}f(x(t),y(t))=\lim_{h->0}\frac{f(x(t)+hn(t),y(t))-f(x(t),y(t))}{h}=\lim_{h->0}\frac{2x(t)hn(t)+h^2n(t)^2}{h}=2x(t)n(t)hence the notation : \frac{df}{dx}=2x(t)Which, in this case
  • #1
adamg
48
0
if you are given f(x,y)=x^2+y^2 and y=cos(t) x=sin(t), then when you differentiate f with respect to t, you use the partial derivatives of f with respect to x and y in the process. When i was taught partial derivatives, i was told that we "keep all but one of the independent variables fixed...". Now in this case, when differentiating f with respect to y, say, i don't see how this works. For, x (=cos(t) ) cannot be fixed while y (=sin(t) ) varies, can it?

When we differentiate f do we 'forget' that x and y are functions of t, and treat them as independent?
 
Mathematics news on Phys.org
  • #2
[tex] f(x,y) = x(t)^2 + y(t)^2 [/tex]

[tex] f(x,y)_x = 2x(t)x'(t) [/tex]

[tex] f(x,y)_y = 2y(t)y'(t) [/tex]


You keep one constant becuse you want to find the rate of change of f(x,y) as x changes, not y. Same thing for finding y.
 
Last edited:
  • #3
When i was taught partial derivatives, i was told that we "keep all but one of the independent variables fixed...". Now in this case, when differentiating f with respect to y, say, i don't see how this works. For, x (=cos(t) ) cannot be fixed while y (=sin(t) ) varies, can it?
The independant variables for F, are x and y. t has no concern here. Why would one not be able to be fixed while the others are moving, this is the exact same thing you do when you do partial derivatives anyway.

When we differentiate f do we 'forget' that x and y are functions of t, and treat them as independent?
Exactly.
 
  • #4
ok, thanks. is the partial derivative of f with respect to x just 2x then?
i don't understand the x'(t) part that you included?
 
  • #5
Its an application of the chain rule, which come to think of it doesn't belong there unless your finding the partial wrt to t.
 
  • #6
I don't know this well, but if you write f(x,y)=x^2+y^2...then [tex] \partial_xf(x,y)=2x[/tex]...?? It's a weird question :

[tex]\frac{d}{dt}(\frac{\partial f}{\partial x}(x,y))=\frac{d2x(t)}{dt}=2x'(t)[/tex]

where as normally, f(x(t),y(t))=g(t) is a function of t only, hence [tex]\partial_xg(t)[/tex]=0...is like if Schwarz's thm were not valid here...

In fact I would say : it depends on at which time you apply the transformation x->x(t) (hence x as variable, or x as function of a variable...)

Because you could see x and y as function : x(t), y(t), and [tex] f(x,y)=x(t)^2+y(t)^2[/tex] as a functional of x and y...

Then you could apply Gateaux derivatives in the "direction" of the function n...(n:t->n(t))...with the usual definition :

[tex] D_{x,n(t)}F(x,y)=\lim_{h->0}\frac{F(x+hn,y)-F(x,y)}{h}=\lim_{h->0}\frac{x(t)^2+2hn(t)x(t)+y(t)^2-x(t)^2-y(t)^2}{h} [/tex]
[tex]=2n(t)x(t) [/tex]

so that we recover whozum result by functionally deriving along n(t)=x'(t)
 
Last edited:
  • #7
whozum said:
The independant variables for F, are x and y. t has no concern here. Why would one not be able to be fixed while the others are moving, this is the exact same thing you do when you do partial derivatives anyway.


.

thats what i was confused about. we treat the x and y as independent variables, even though they won't really vary independently (since both depend on t)
 
  • #8
So what is your conclusion klein? What is

[tex] \frac{\delta f}{\delta x} \ and \ \frac{\delta f}{\delta t} [/tex] ?
 
  • #9
That's the notation for the functional derivative of "f" wrt "x" or "t".

Daniel.
 
  • #10
whozum said:
[tex] f(x,y) = x(t)^2 + y(t)^2 [/tex]

[tex] f(x,y)_x = 2x(t)x'(t) [/tex]

[tex] f(x,y)_y = 2y(t)y'(t) [/tex]


You keep one constant becuse you want to find the rate of change of f(x,y) as x changes, not y. Same thing for finding y.

This makes no sense at all: x(t)2+ y(t)2 is a function of t, not x and y, and so cannot be equal to f(x,y).
Even worse is [tex] f(x,y)_x = 2x(t)x'(t) [/tex] and [tex] f(x,y)_y = 2y(t)y'(t) [/tex]. If f is a function of t, then it makes no sense to write "fx(x,y).

What IS true is that if f(x,y)= x2+ y2, then fx(x,y)= 2x and fy(x,y)= 2y.
IF, further, x and y are themselves functions of t, then f is really a function of t, f(t), and f '(t)= (2x)x'(t)+ (2y)y'(t).
 
  • #11
what about if you had f(x,y) = x^2+y^2 again, but with y a function of x. The for the partial derivative of f with respect to x, you keep y fixed (?) and let x vary, even though y is a function of x (?). Thanks.
 
  • #12
IF, further, x and y are themselves functions of t, then f is really a function of t, f(t), and f '(t)= (2x)x'(t)+ (2y)y'(t).

So [itex] f_x(x,y) [/itex] doesn't really exist? What you described is what I did but just assuming that you could differentiate withrespect to each variable.
 
  • #13
What whozum did is the following :

[tex]f(x,y)=x(t)^2+y(t)^2[/tex]...(which by the way is a functional but not a function)

Then you thought, ok I want to differentiate with respect to x...hence I should move x a bit..the problem is that x is a function, hence I don't vary with another function, but the parameter of this function : t

so you did :

[tex] \lim_{h->0}\frac{f(x(t+h),y(t))-f(x(t),y(t))}{h}=\lim_{h->0}\frac{x(t+h)^2-x(t)^2}{h}\approx\lim_{h->0}\frac{(x(t)+hx'(t)+...)^2-x(t)^2}{h}=2x(t)x'(t)[/tex]

But you could eventually vary x with a function n, instead of varying the variable of x...
 
  • #14
Somehow, I doubt whozum was thinking along the lines of functionals here.
whozum:
Let x,y be given as functions x=X(t), y=Y(t)
What is now correct is that we may define a function F(t)=f(X(t),Y(t))
Then, we have by the chain rule:
[tex]\frac{dF}{dt}=\frac{\partial{f}}{\partial{x}}\mid_{(x,y)=(X(t),Y(t))}}\frac{dX}{dt}+\frac{\partial{f}}{\partial{y}}\mid_{(x,y)=(X(t),Y(t))}}\frac{dY}{dt}[/tex]
As HallsofIvy has already said.
 
Last edited:
  • #15
arildno said:
Somehow, I doubt whozum was thinking along the lines of functionals here.
whozum:
Let x,y be given as functions x=X(t), y=Y(t)
What is now correct is that we may define a function F(t)=f(X(t),Y(t))
Then, we have by the chain rule:
[tex]\frac{dF}{dt}=\frac{\partial{f}}{\partial{x}}\mid_{(x,y)=(X(t),Y(t))}}\frac{dX}{dt}+\frac{\partial{f}}{\partial{y}}\mid_{(x,y)=(X(t),Y(t))}}\frac{dY}{dt}[/tex]
As HallsofIvy has already said.

Thanks for doubting me ;) Its justified though.

You are correct, and I understand what HallsofIvy said and what you are saying, but the question presnted by the OP was the last question I posted two/three posts ago, this is the only question I have left.

I'm also trying to think of a similar instance where the indep. variable is a function of another variable, I will let you know.

Thanks.
 

1. What is a partial derivative?

A partial derivative is a mathematical concept used to measure the rate of change of a function with respect to one of its variables, while holding all other variables constant. It is represented by the symbol ∂ (pronounced "partial").

2. How is a partial derivative different from a regular derivative?

A regular derivative measures the rate of change of a function with respect to a single variable, while a partial derivative measures the rate of change of a multivariable function with respect to one of its variables, while keeping all other variables constant.

3. What is the purpose of finding partial derivatives?

Partial derivatives are used in many areas of science and mathematics, including physics, economics, and engineering. They are particularly useful in optimization problems, where we need to find the maximum or minimum value of a function with multiple variables.

4. How do you calculate a partial derivative?

To calculate a partial derivative, you can use the same rules and techniques as for regular derivatives, except you only need to differentiate with respect to the variable in question, treating all other variables as constants. This means you can use the power rule, product rule, quotient rule, and chain rule as needed.

5. Are there any practical applications of partial derivatives?

Yes, partial derivatives have many practical applications in fields such as physics, engineering, economics, and statistics. For example, in physics, we can use partial derivatives to calculate the velocity and acceleration of an object moving in multiple dimensions. In economics, they can be used to determine the marginal rate of substitution and elasticity of demand. In statistics, they are used in regression analysis and to calculate the sensitivity of a model to various inputs.

Similar threads

Replies
2
Views
126
  • General Math
Replies
5
Views
308
Replies
3
Views
575
Replies
1
Views
796
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
505
Replies
1
Views
93
Replies
2
Views
1K
  • General Math
Replies
11
Views
1K
Replies
1
Views
1K
Back
Top