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Partial derivatives function

  1. Nov 20, 2013 #1
    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...
     
  2. jcsd
  3. Nov 21, 2013 #2

    Mark44

    Staff: Mentor

    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.
    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.
    For most functions that you're likely to come across, the mixed partials, as they're called, are the same.
     
  4. Nov 21, 2013 #3
    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?
     
  5. Nov 21, 2013 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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)?
     
  6. Nov 21, 2013 #5
    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 ...
     
  7. Nov 23, 2013 #6
    Someone have more some opinion?
     
  8. Nov 23, 2013 #7
    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.

    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.
     
  9. Nov 23, 2013 #8
    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.
     
  10. Nov 25, 2013 #9

    Mark44

    Staff: Mentor

    By breaking it down to what it means; namely, ∂/∂t(∂f/∂s).
    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.
    Mandelbroth answered your question, or at least pointed you to some sources for answers.
     
  11. Nov 25, 2013 #10
    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]
     
  12. Nov 26, 2013 #11
    Chestermiller, your comment is very interesting to me! How did you come to this two additional terms?
     
  13. Nov 26, 2013 #12
    Just differentiation by parts:
    $$\frac{\partial}{\partial s} ({\frac{\partial f}{\partial x} \frac{\partial x}{\partial t}})=\left(\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial s}\frac{\partial x}{\partial t}+\frac{\partial^2 f}{\partial x \partial y}\frac{\partial y}{\partial s}\frac{\partial x}{\partial t}\right)+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial s\partial t}$$
     
  14. Nov 26, 2013 #13

    Mark44

    Staff: Mentor

    I've never heard of "differentiation by parts." (Integration by parts, yes.) Aren't you just using the product and chain rules?
     
  15. Nov 26, 2013 #14
    With relation to this, I don't know how appeared the ∂y above!?
     
  16. Nov 26, 2013 #15
    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. Nov 26, 2013 #16
    f is a function of both x and y, and y is a function of s.
     
  18. Nov 26, 2013 #17
    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).
     
  19. Nov 26, 2013 #18
    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.
     
  20. Nov 26, 2013 #19
    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]
     
  21. Nov 26, 2013 #20
    Yes. If f is a function only of x, and not y, then this, of course, is correct.
     
    Last edited by a moderator: Nov 26, 2013
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