Partial derivatives function

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  • #1
Jhenrique
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Se a function f(x(t, s), y(t, s)) have as derivative with respect to t:

[tex]\frac{df}{dt}=\frac{df}{dx} \frac{dx}{dt}+\frac{df}{dy} \frac{dy}{dt}[/tex]

And, with respect to s:

[tex]\frac{df}{ds}=\frac{df}{dx} \frac{dx}{ds}+\frac{df}{dy} \frac{dy}{ds}[/tex]

But, how will be the derivative with respect to t and s?

[tex]\frac{d^2f}{dtds}[/tex]

Or with respect to s and t

[tex]\frac{d^2f}{dsdt}[/tex]

I don't know if there is difference when change the order between t and s...
 

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  • #2
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Se a function f(x(t, s), y(t, s)) have as derivative with respect to t:

[tex]\frac{df}{dt}=\frac{df}{dx} \frac{dx}{dt}+\frac{df}{dy} \frac{dy}{dt}[/tex]

And, with respect to s:

[tex]\frac{df}{ds}=\frac{df}{dx} \frac{dx}{ds}+\frac{df}{dy} \frac{dy}{ds}[/tex]
Since f is ultimately a function of two variables, t and s, it doesn't make sense to talk about df/dt or df/ds, the derivatives of f with respect to t and s.

Instead we can talk about the partial derivative of f with respect to t or to s, using this notation for your first equation above:
$$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x} \frac{\partial x}{\partial t}+\frac{\partial f}{\partial y} \frac{\partial y}{\partial t} $$

There are other forms of notation in use, such as fx to indicate the partial derivative (or just partial) of f with respect to x.
But, how will be the derivative with respect to t and s?

[tex]\frac{d^2f}{dtds}[/tex]
We write that as
$$\frac{\partial^2 f}{\partial t \partial s} $$

This is calculated as
$$ \frac{\partial}{\partial t}\left( \frac{\partial f}{\partial s}\right)$$

In other words, take the partial of f with respect to s, and then take the partial of that with respect to t.
Or with respect to s and t

[tex]\frac{d^2f}{dsdt}[/tex]

I don't know if there is difference when change the order between t and s...
For most functions that you're likely to come across, the mixed partials, as they're called, are the same.
 
  • #3
Jhenrique
685
4
I think that this
[tex]\frac{\partial^2 f}{\partial s \partial t} = \frac{\partial^2 f}{\partial x^2} \frac{dxdx}{\partial s \partial t} + 2 \frac{\partial^2 f}{\partial x \partial y} \frac{dxdy}{\partial s \partial t} + \frac{\partial^2 f}{\partial y^2} \frac{dydy}{\partial s \partial t}[/tex]
is different this
[tex]\frac{\partial^2 f}{\partial t \partial s} = \frac{\partial^2 f}{\partial x^2} \frac{dxdx}{\partial t \partial s} + 2 \frac{\partial^2 f}{\partial x \partial y} \frac{dxdy}{\partial t \partial s} + \frac{\partial^2 f}{\partial y^2} \frac{dydy}{\partial t \partial s}[/tex]
But, when wrote this form
[tex]\frac{\partial^2 f}{\partial s \partial t} = \frac{\partial^2 f}{\partial x^2} \frac{dxdx}{\partial s \partial t} + \frac{\partial^2 f}{\partial x \partial y} \frac{dxdy}{\partial s \partial t} + \frac{\partial^2 f}{\partial y \partial x} \frac{dydx}{\partial s \partial t} + \frac{\partial^2 f}{\partial y^2} \frac{dydy}{\partial s \partial t}[/tex]
it is not different this
[tex]\frac{\partial^2 f}{\partial t \partial s} = \frac{\partial^2 f}{\partial x^2} \frac{dxdx}{\partial t \partial s} + \frac{\partial^2 f}{\partial x \partial y} \frac{dxdy}{\partial t \partial s} + \frac{\partial^2 f}{\partial y \partial x} \frac{dydx}{\partial t \partial s} + \frac{\partial^2 f}{\partial y^2} \frac{dydy}{\partial t \partial s}[/tex]
I think...

I still wonder what comes to be?
[tex]\frac{dxdy}{dtds}[/tex]
Would be a Jacobian? Would be a product between dx/dt and dy/ds?
 
  • #4
HallsofIvy
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Where did you see that notation? I would recognise [itex]\dfrac{dx}{dt}\frac{dy}{ds}[/itex] as a product of derivatives and [itex]\dfrac{d(xy)}{d(st)}[/itex] as a Jacobian but you seem to have a mixture of those notations with only a single "vinculum" indicating a single fraction but 2 "d"s in both numerator and denominator. Wht information do you have about x and y? Is x a function of t only and y a function of s only (most likely badly written product of derivatives) or are x and y each functions of both t and s (most likely Jacobian)?
 
  • #5
Jhenrique
685
4
I posted the function in the first post. f(x(t, s), y(t, s)). And I from the relationship:
[tex]d^2f=(\frac{\partial }{\partial x} dx + \frac{\partial }{\partial y} dy)^2f[/tex]
Adding dtds in denominator. But I don't know do this with all mathematical rigor. I have my doubts ...
 
  • #6
Jhenrique
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Someone have more some opinion?
 
  • #7
Mandelbroth
611
24
Se a function f(x(t, s), y(t, s)) have as derivative with respect to t:

[tex]\frac{df}{dt}=\frac{df}{dx} \frac{dx}{dt}+\frac{df}{dy} \frac{dy}{dt}[/tex]

And, with respect to s:

[tex]\frac{df}{ds}=\frac{df}{dx} \frac{dx}{ds}+\frac{df}{dy} \frac{dy}{ds}[/tex]

But, how will be the derivative with respect to t and s?

[tex]\frac{d^2f}{dtds}[/tex]

Or with respect to s and t

[tex]\frac{d^2f}{dsdt}[/tex]

I don't know if there is difference when change the order between t and s...
Well, aside from the notational problem mentioned by others (you need to use partials), you have come across a fundamental question of multivariate calculus: when can we change the order of partial differentiation?

You might consider looking up Clairaut's Theorem. This theorem is the explanation. The problem is also intimately related to the Poincaré Lemma.

I posted the function in the first post. f(x(t, s), y(t, s)). And I from the relationship:
[tex]d^2f=(\frac{\partial }{\partial x} dx + \frac{\partial }{\partial y} dy)^2f[/tex]
Adding dtds in denominator. But I don't know do this with all mathematical rigor. I have my doubts ...
You are right to have doubts. The exterior derivative (that "differential-making thingy"), denoted ##\mathrm{d}##, is described by several axioms, one of which is ##\mathrm{d}(\mathrm{d}\alpha)=\mathrm{d}^2\alpha=0##, where I use 0 to mean the additive identity for differential forms and ##\alpha## is any form. Your left hand side is 0 and your right hand side means nothing.
 
  • #8
Jhenrique
685
4
I think I messed up the topic title. Should be: "Second derivative parametrized". And I flinched 'cause I didn't used the notation ∂ correctly. But, with all respect, my question still remain! How to derive ∂²f/∂t∂s!? I can also apply the rule chair: ∂/∂s(∂f/∂t), but the result is quite different of equations (1), (2), (3) and (4) at post #3.
 
  • #9
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I think I messed up the topic title. Should be: "Second derivative parametrized". And I flinched 'cause I didn't used the notation ∂ correctly. But, with all respect, my question still remain! How to derive ∂²f/∂t∂s!?
By breaking it down to what it means; namely, ∂/∂t(∂f/∂s).
I can also apply the rule chair: ∂/∂s(∂f/∂t)
That's not the chain rule. ∂²f/∂t∂s means ∂/∂t(∂f/∂s). What you have, ∂/∂s(∂f/∂t), means taking the partial derivatives in the reverse order.
, but the result is quite different of equations (1), (2), (3) and (4) at post #3.

Mandelbroth answered your question, or at least pointed you to some sources for answers.
Mandelbroth said:
Well, aside from the notational problem mentioned by others (you need to use partials), you have come across a fundamental question of multivariate calculus: when can we change the order of partial differentiation?

You might consider looking up Clairaut's Theorem. This theorem is the explanation. The problem is also intimately related to the Poincaré Lemma.
 
  • #10
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[tex]\frac{\partial^2 f}{\partial s \partial t} = \frac{\partial^2 f}{\partial x^2} \frac{dxdx}{\partial s \partial t} + \frac{\partial^2 f}{\partial x \partial y} \frac{dxdy}{\partial s \partial t} + \frac{\partial^2 f}{\partial y \partial x} \frac{dydx}{\partial s \partial t} + \frac{\partial^2 f}{\partial y^2} \frac{dydy}{\partial s \partial t}[/tex]

[tex]\frac{\partial^2 f}{\partial t \partial s} = \frac{\partial^2 f}{\partial x^2} \frac{dxdx}{\partial t \partial s} + \frac{\partial^2 f}{\partial x \partial y} \frac{dxdy}{\partial t \partial s} + \frac{\partial^2 f}{\partial y \partial x} \frac{dydx}{\partial t \partial s} + \frac{\partial^2 f}{\partial y^2} \frac{dydy}{\partial t \partial s}[/tex]
I got something similar to this, but I got two additional terms:
[tex]\frac{\partial^2 f}{\partial s \partial t} =\frac{\partial^2 f}{\partial t \partial s}= \frac{\partial^2 f}{\partial x^2} \frac{\partial x}{\partial s }\frac{\partial x}{\partial t } + \frac{\partial^2 f}{\partial x \partial y}\left( \frac{\partial y}{\partial s }\frac{\partial x}{\partial t } + \frac{\partial x}{\partial s }\frac{\partial y}{\partial t }\right) + \frac{\partial^2 f}{\partial y^2} \frac{\partial y}{\partial s }\frac{\partial y}{\partial t }+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial s \partial t}+\frac{\partial f}{\partial y}\frac{\partial^2 y}{\partial s \partial t}[/tex]
 
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  • #11
Jhenrique
685
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I got something similar to this, but I got two additional terms:
[tex]\frac{\partial^2 f}{\partial s \partial t} =\frac{\partial^2 f}{\partial t \partial s}= \frac{\partial^2 f}{\partial x^2} \frac{\partial x}{\partial s }\frac{\partial x}{\partial t } + \frac{\partial^2 f}{\partial x \partial y}\left( \frac{\partial y}{\partial s }\frac{\partial x}{\partial t } + \frac{\partial x}{\partial s }\frac{\partial y}{\partial t }\right) + \frac{\partial^2 f}{\partial y^2} \frac{\partial y}{\partial s }\frac{\partial y}{\partial t }+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial s \partial t}+\frac{\partial f}{\partial y}\frac{\partial^2 y}{\partial s \partial t}[/tex]

Chestermiller, your comment is very interesting to me! How did you come to this two additional terms?
 
  • #12
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Chestermiller, your comment is very interesting to me! How did you come to this two additional terms?
Just differentiation by parts:
$$\frac{\partial}{\partial s} ({\frac{\partial f}{\partial x} \frac{\partial x}{\partial t}})=\left(\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial s}\frac{\partial x}{\partial t}+\frac{\partial^2 f}{\partial x \partial y}\frac{\partial y}{\partial s}\frac{\partial x}{\partial t}\right)+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial s\partial t}$$
 
  • #13
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Just differentiation by parts:
$$\frac{\partial}{\partial s} ({\frac{\partial f}{\partial x} \frac{\partial x}{\partial t}})=\left(\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial s}\frac{\partial x}{\partial t}+\frac{\partial^2 f}{\partial x \partial y}\frac{\partial y}{\partial s}\frac{\partial x}{\partial t}\right)+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial s\partial t}$$
I've never heard of "differentiation by parts." (Integration by parts, yes.) Aren't you just using the product and chain rules?
 
  • #14
Jhenrique
685
4
$$\frac{\partial}{\partial s} ({\frac{\partial f}{\partial x} \frac{\partial x}{\partial t}})=\left(\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial s}\frac{\partial x}{\partial t}+\frac{\partial^2 f}{\partial x \partial y}\frac{\partial y}{\partial s}\frac{\partial x}{\partial t}\right)+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial s\partial t}$$

With relation to this, I don't know how appeared the ∂y above!?
 
  • #15
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I've never heard of "differentiation by parts." (Integration by parts, yes.) Aren't you just using the product and chain rules?
Yes. You're right. For some reason, I always call it differentiation by parts, but it's really the product rule. I can't remember when I started using this unusual terminology.
 
  • #17
Jhenrique
685
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f is a function of both x and y, and y is a function of s.

but y is function of t too! and y isn't in the left side of equation where you derivative with relation to t. Or f is function of x(t, s) and y(t, s) or f is only function of x(t, s).
 
  • #18
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but y is function of t too! and y isn't in the left side of equation where you derivative with relation to t. Or f is function of x(t, s) and y(t, s) or f is only function of x(t, s).
I don't understand what you are asking. f is a function of x and y, so all its partial derivatives with respect to either x , y, or both are also functions of x and y. And x and y are functions of t and s.
 
  • #19
Jhenrique
685
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I want say to f(x(t, s), y(t, s)), ∂²f/∂s∂t is
[tex]\frac{\partial}{\partial s} \left({\frac{\partial f}{\partial x} \frac{\partial x}{\partial t}} + {\frac{\partial f}{\partial y} \frac{\partial y}{\partial t}} \right)=\frac{\partial^2 f}{\partial x^2} \frac{\partial x}{\partial s }\frac{\partial x}{\partial t } + \frac{\partial^2 f}{\partial x \partial y}\left( \frac{\partial y}{\partial s }\frac{\partial x}{\partial t } + \frac{\partial x}{\partial s }\frac{\partial y}{\partial t }\right) + \frac{\partial^2 f}{\partial y^2} \frac{\partial y}{\partial s }\frac{\partial y}{\partial t }+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial s \partial t}+\frac{\partial f}{\partial y}\frac{\partial^2 y}{\partial s \partial t}[/tex]
And to f(x(t, s)), ∂²f/∂s∂t should be
[tex]\frac{\partial}{\partial s} \left({\frac{\partial f}{\partial x} \frac{\partial x}{\partial t}}\right)=\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial s}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial s\partial t}[/tex]
 
  • #20
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I want say to f(x(t, s), y(t, s)), ∂²f/∂s∂t is
[tex]\frac{\partial}{\partial s} \left({\frac{\partial f}{\partial x} \frac{\partial x}{\partial t}} + {\frac{\partial f}{\partial y} \frac{\partial y}{\partial t}} \right)[/tex]
[tex]=\frac{\partial^2 f}{\partial x^2} \frac{\partial x}{\partial s }\frac{\partial x}{\partial t } + \frac{\partial^2 f}{\partial x \partial y}\left( \frac{\partial y}{\partial s }\frac{\partial x}{\partial t } + \frac{\partial x}{\partial s }\frac{\partial y}{\partial t }\right) + \frac{\partial^2 f}{\partial y^2} \frac{\partial y}{\partial s }\frac{\partial y}{\partial t }+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial s \partial t}+\frac{\partial f}{\partial y}\frac{\partial^2 y}{\partial s \partial t}[/tex]
And to f(x(t, s)), ∂²f/∂s∂t should be
[tex]\frac{\partial}{\partial s} \left({\frac{\partial f}{\partial x} \frac{\partial x}{\partial t}}\right)=\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial s}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial s\partial t}[/tex]
Yes. If f is a function only of x, and not y, then this, of course, is correct.
 
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  • #21
Jhenrique
685
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So, how contrast these 2 equations apparently unequal!?

Just differentiation by parts:
$$\frac{\partial}{\partial s} ({\frac{\partial f}{\partial x} \frac{\partial x}{\partial t}})=\left(\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial s}\frac{\partial x}{\partial t}+\frac{\partial^2 f}{\partial x \partial y}\frac{\partial y}{\partial s}\frac{\partial x}{\partial t}\right)+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial s\partial t}$$
And to f(x(t, s)), ∂²f/∂s∂t should be
[tex]\frac{\partial}{\partial s} \left({\frac{\partial f}{\partial x} \frac{\partial x}{\partial t}}\right)=\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial s}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial s\partial t}[/tex]
 
  • #22
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So, how contrast these 2 equations apparently unequal!?
The first one assumes that f depends on y, and the second one assumes that f does not depend on y. So, of course, you would expect to get two different results. Why is this surprising to you?
 
  • #23
Jhenrique
685
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oooh, I understood! thx!
 
  • #24
Jhenrique
685
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NO, wait! Supposing f(x(t, s), y(t, s)), the equation correct isn't this????
[tex]\frac{\partial }{\partial s}\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial t} \right )[/tex]
[tex]=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial t} \right )\frac{\partial x}{\partial s}+\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial t} \right )\frac{\partial y}{\partial s}=\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial t}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial x \partial t}\frac{\partial x}{\partial s} + \frac{\partial^2 f}{\partial y \partial x}\frac{\partial x}{\partial t}\frac{\partial y}{\partial s}+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial y \partial t}\frac{\partial y}{\partial s}[/tex]
 
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  • #25
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NO, wait! Supposing f(x(t, s), y(t, s)), the equation correct isn't this????
[tex]\frac{\partial }{\partial s}\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial t} \right )[/tex]
[tex]=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial t} \right )\frac{\partial x}{\partial s}+\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial t} \right )\frac{\partial y}{\partial s}=\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial t}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial x \partial t}\frac{\partial x}{\partial s} + \frac{\partial^2 f}{\partial y \partial x}\frac{\partial x}{\partial t}\frac{\partial y}{\partial s}+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial y \partial t}\frac{\partial y}{\partial s}[/tex]
What the heck is [itex]\frac{\partial^2 x}{\partial x \partial t}[/itex] supposed to mean? What's being held constant? Still, if you want to persist with this formalism, then if you write:
[tex]\frac{\partial }{\partial s}(\frac{\partial x}{\partial t})=\frac{\partial^2 x}{\partial x \partial t}\frac{\partial x}{\partial s} +\frac{\partial^2 x}{\partial y \partial t}\frac{\partial y}{\partial s}[/tex]
then you get:
[tex]\frac{\partial^2 x}{\partial x \partial t}\frac{\partial x}{\partial s} +\frac{\partial^2 x}{\partial y \partial t}\frac{\partial y}{\partial s}=\frac{\partial^2 x}{\partial s\partial t}[/tex]
That gets you back to the same result I gave.
 
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  • #26
Jhenrique
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That's complicated!!! LOOOL thx!
 

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