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Difference between d²y/d²x and d²y/dx²

  1. Dec 27, 2007 #1
    Considering a parametric function:

    [tex]
    y = sin(\theta)
    [/tex]
    [tex]
    x = cos(\theta)
    [/tex]

    I want to find it's second order derivative.

    [tex]
    \frac{dy}{d\theta} = cos(\theta)
    [/tex]

    [tex]
    \frac{dx}{d\theta} = -sin(\theta)
    [/tex]

    Therefore, I can say that:

    [tex]
    \frac{dy}{dx} = -cot(\theta)
    [/tex]

    Also,

    [tex]
    \frac{d^2y}{d\theta^2} = -sin(\theta)
    [/tex]

    [tex]
    \frac{d^2x}{d\theta^2} = -cos(\theta)
    [/tex]

    Hence,

    [tex]
    \frac{d^2y}{d^2x} = tan(\theta)
    [/tex]

    But, If i need the second order derivate, what i actually need is:

    [tex]
    \frac{d^2y}{dx^2} = \frac{d}{d\theta}(\frac{dy}{dx})\times(\frac{d\theta}{dx})
    [/tex]

    which turns out to be:

    [tex]
    \frac{d^2y}{dx^2} = -cosec^3(\theta)
    [/tex]

    So, basically, what is the difference between d²y/d²x and d²y/dx². And how can we arrive at the formula for d²y/dx²? I mean, what is the mathematical significance of that? And how to do it for higher order derivatives, let's say 3 or 4 for example.

    Also, does something like d²y/d²x even exist? what does it mean?
     
    Last edited: Dec 28, 2007
  2. jcsd
  3. Dec 27, 2007 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    No, it does not exist. We specifically use the notation d2y/dx2 to indicate that something more than mere division (as you did in the original post) is taking place.
     
  4. Dec 27, 2007 #3

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Are you sure that's right? I think I get a different answer using y =\sqrt{1-x^2}...
     
  5. Dec 27, 2007 #4
    I'll explain two things. First how to do this

    more rigorously, and then why I think that this

    is wrong.

    If we first have functions

    [tex]
    \theta\mapsto y(\theta)
    [/tex]
    [tex]
    \theta\mapsto x(\theta)
    [/tex]

    and you want to reparametrize y as a function of x, to get some function [itex]x\mapsto y[/itex], strictly speaking this is going to be different function y, so it would be clearer to give it a different notation. I'll put [tex]\tilde{y}[/tex]. We must assume that we can invert the mapping [itex]\theta\mapsto x(\theta)[/itex] to a mapping [itex]x\mapsto \theta(x)[/itex], and then y and y-tilde are related by

    [tex]
    \tilde{y}(x) = y(\theta(x))
    [/tex]

    The derivative is then

    [tex]
    \frac{d\tilde{y}(x)}{dx} = D_x y(\theta(x)) = \frac{dy(\theta(x))}{d\theta} \frac{d\theta(x)}{dx},
    [/tex]

    or

    [tex]
    \frac{dy}{d\theta}\frac{d\theta}{dx}
    [/tex]

    for short, without parameters. In your example we must assume (for example) [itex]\theta > 0[/itex] so that [itex]\theta(x)=\cos^{-1}(x)[/itex] exists. Then we can substitute

    [tex]
    \frac{dy}{d\theta} = \cos\theta
    [/tex]

    [tex]
    \frac{d\theta}{dx} = -\frac{1}{\sqrt{1-x^2}} = -\frac{1}{\sin\theta}
    [/tex]

    and get

    [tex]
    \frac{d\tilde{y}}{dx} = -\frac{\cos\theta}{\sin\theta} = -\cot\theta
    [/tex]

    You got this same result more easily, but check what the second derivative turns out to be.

    [tex]
    \frac{d^2\tilde{y}(x)}{dx^2} = D_x\Big(\frac{dy(\theta(x))}{d\theta} \frac{d\theta(x)}{dx}\Big) = \Big(D_x \frac{dy(\theta(x))}{d\theta}\Big) \frac{d\theta(x)}{dx}\; + \;\frac{dy(\theta(x))}{d\theta}\Big( D_x \frac{d\theta(x)}{dx}\Big)
    [/tex]

    [tex]
    = \Big(\frac{d^2 y(\theta(x))}{d\theta^2} \frac{d\theta(x)}{dx}\Big) \frac{d\theta(x)}{dx}\; +\; \frac{dy(\theta(x))}{d\theta} \frac{d^2\theta(x)}{dx^2} = \frac{d^2 y(\theta(x))}{d\theta^2} \Big(\frac{d\theta(x)}{dx}\Big)^2\; +\; \frac{dy(\theta(x))}{d\theta} \frac{d^2\theta(x)}{dx^2}
    [/tex]

    or

    [tex]
    \frac{d^2y}{d\theta^2} \Big(\frac{d\theta}{dx}\Big)^2 + \frac{dy}{d\theta} \frac{d^2\theta}{dx^2}
    [/tex]

    for short, without parameters. By substituting

    [tex]
    \frac{d^2y}{d\theta} = -\sin\theta
    [/tex]

    [tex]
    \frac{d^2\theta}{dx^2} = -\frac{x}{(1-x^2)^{3/2}} = -\frac{\cos\theta}{\sin^3\theta}
    [/tex]

    and some previous expressions, we get

    [tex]
    \frac{d^2\tilde{y}}{dx^2} = -\sin\theta \frac{1}{\sin^2\theta}\; +\; \cos\theta\frac{-\cos\theta}{\sin^3\theta} = -\frac{1}{\sin^3\theta} = -\textrm{cosec}^3\theta
    [/tex]
     
    Last edited: Dec 28, 2007
  6. Dec 27, 2007 #5
    Hehe. Or alternatively one can calculate

    [tex]
    D^2_x \sqrt{1-x^2} = D_x \frac{-x}{\sqrt{1-x^2}} = \cdots = -\frac{1}{(1-x^2)^{3/2}}
    [/tex]

    :redface:

    well, rohanprabhu, I hope you find the equation

    [tex]
    \frac{d^2y}{dx^2} = \frac{d^2y}{d\theta^2} \Big(\frac{d\theta}{dx}\Big)^2 + \frac{dy}{d\theta} \frac{d^2\theta}{dx^2}
    [/tex]

    interesting, at least. It could be useful if the mappings [itex]\theta\mapsto y(\theta)[/itex] and [itex]\theta\mapsto x(\theta)[/itex] were something different.
     
    Last edited: Dec 28, 2007
  7. Dec 28, 2007 #6
    i'm not sure that this is right or not.. but i've been using this formula for my textbook questions.. and it sorta works. But since my textbooks doesn't really have advanced problems to deal with, i am not sure how applicable this formula is..

    EDIT1:

    no way.. i am totally wrong in this one. I tried doing using

    [tex]
    y = \sqrt{1 - x^2}
    [/tex]

    and i'm getting a totally different answer. I need to refer some valid source first..

    EDIT2:

    in the second method, I multiplied the values wrongly. I am not getting cosec(θ), but cosec³θ . But still, i'm losing a (-) sign somewhere.

    @jostpuur: the post u made was p.e.r.f.e.c.t. Beautiful explaination.. thanks a lot.. now i truly understand it. and also.. ur formula was perfect...

    P.S: you seriously typed all that in LaTeX?
    P.S2: Could u tell me what exactly θ → x(θ) means.. does it mean something like, for a set of values of 'θ', we have a set of values of 'x' or that the argument of the function y-tilde varies as a function of 'θ'.
     
    Last edited: Dec 28, 2007
  8. Dec 28, 2007 #7
    I later realized there's a chance, that I just didn't understand what

    was supposed to mean, and that you had done a calculation mistake after this. Could you explain what this is?

    For example, if you want to define a mapping [itex]f:\mathbb{R}\to\mathbb{R}[/itex], [itex]f(x)=x^2[/itex], you could alternatively do it by talking about mapping [itex]\mathbb{R}\to\mathbb{R}[/itex], [itex]x\mapsto x^2[/itex]. The arrow used to define how each member gets mapped is \mapsto, which is different from the usual one.
     
  9. Dec 28, 2007 #8
    This is no-life.
     
  10. Dec 28, 2007 #9
    This was a formula taught to us for the 2nd order derivative of a parametric function.

    totally ;)
     
    Last edited: Dec 28, 2007
  11. Dec 28, 2007 #10
    Eureka.. the formula you provided and the one taught to me by my teacher is the same.. I spent some time over this, and arrived at this proof [finally :D]..

    I have attached a screenshot of the proof, as I did the typeset in Mathematica [typing that much LaTeX would have been a pain...]
     

    Attached Files:

    Last edited: Dec 28, 2007
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