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Chain Rule, Differentials "Trick"

  1. Aug 23, 2014 #1
    I was playing around with some simple differential equations earlier and I discovered something cool (at least for me).

    Suppose you have [tex]y=sin(x^2) \Rightarrow \frac{dy}{dx}=2xcos(x^2)[/tex]
    What if, instead of taking the derivative with respect to [itex]x[/itex], I want to take the derivative with respect to [itex]x^2[/itex], where a simple substitution might help? [tex]u(x)=x^2 \Rightarrow \frac{du}{dx}=2x \Rightarrow du=2xdx[/tex] [tex]y=sin(u) \Rightarrow \frac{dy}{du}=cos(u) \Rightarrow \frac{dy}{dx^2}=cos(x^2)[/tex] If you undo the substitution for both instances of [itex]u[/itex], you arrive at the derivative with respect to [itex]x[/itex] [tex]\frac{dy}{dx^2}=\frac{dy}{2xdx}=cos(u^2) \Rightarrow \frac{dy}{dx}=2xcos(x^2)[/tex]

    Now, I understand this result is just an application of the chain rule, but, I decided to use it on integrals. Which turns out great because this is just an application of integration by parts. [tex]\int dy=\int 2xcos(x^2)dx=\int cos(u)du=\int cos(x^2)dx^2[/tex]

    Now, my question is, is it possible to apply this to derivatives of higher order? Here is my approach by example. [tex]\frac{d^2y}{dx^2}=6x[/tex] [tex]\int d^2y=\int 6xdx^2,~~~~~~~dx^2=2xdx[/tex] [tex]\int d^2y=\int 6x(2xdx)[/tex]

    In analogy to stating [itex]\int dx^2=\int 2xdx[/itex], how would i write [itex]\int d^2y ~~~~[/itex] ?
     
  2. jcsd
  3. Aug 23, 2014 #2

    pwsnafu

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    It's an application of the Riemann-Stieltjes integral.
    It's not really a differential thing.
     
  4. Aug 25, 2014 #3
    I read the article to the best of my understanding and I didn't notice anything that could help me write [itex]\int d^2y[/itex] explicitly so I could compute the integral using only one step of anti-differentiation (rather than two, in the case of second-order derivatives).
     
    Last edited: Aug 25, 2014
  5. Aug 25, 2014 #4
    The trick that they teach you in freshman calculus and then again in DiffEQ where you take a separable DE ##\frac{dy}{dx}=f(x)g(y)## and treat ##\frac{dy}{dx}## like a ratio of differentials so that the equation can be rewritten ##\frac{dy}{g(y)}=f(x)dx## is an abuse of notation. It works (in the sense that you get a correct answer) in that particular situation, and in some others. But in general, treating the Leibniz notation like it's a fraction will get you in trouble if you don't really know what you're doing.

    In the case of the second derivative, ##\frac{d^2y}{dx^2}##, it most definitely cannot be treated like a fraction. Furthermore, you have conflated ##d(x^2)## with what "should" be (if it were actually a valid step) ##(dx)^2##.

    For what it's worth, here is the legitimate working of the general separable DE;

    Given ##\frac{dy}{dx}=f(x)g(y)## has a solution ##y=u(x)##, we have ##u'(x)=f(x)g\left(u(x)\right)##. Whenever ##g## is nonzero, ##\frac{1}{g\left(u(x)\right)}u'(x)=f(x)##. On any interval ##[x_0,x_1]## on which the solution is valid, we have $$\int_{x_0}^{x_1}\frac{1}{g\left(u(x)\right)}u'(x)\ dx = \int_{x_0}^{x_1}f(x)\ dx$$ Using the method of integration by substitution on the LHS gives $$\int_{w_0}^{w_1}\frac{1}{g\left(w\right)}\ dw = \int_{x_0}^{x_1}f(x)\ dx$$ where ##w_i=u(x_i)## for ##i=1,2##.

    And this looks a lot like the equation ##\int\frac{1}{g(y)}\ dy=\int f(x)\ dx## that you get using the standard trick. So treating ##\frac{dy}{dx}## like a fraction works in this case.

    If you were to try and apply this approach to a second order separable DE (which is a misnomer) like ##\frac{d^2y}{dx^2}=f(x)g(y)##, you'd get to the equation $$\int_{x_0}^{x_1}\frac{1}{g\left(u(x)\right)}u''(x)\ dx = \int_{x_0}^{x_1}f(x)\ dx$$ and (hopefully) realize that the method of substitution does not apply here.

    Also, the Leibniz notation ##\frac{d^ny}{dx^n}## comes from the notation ##\frac{d}{dx}## of the differential operator (http://en.wikipedia.org/wiki/Differential_operator). ##\frac{d^n}{dx^n}=(\frac{d}{dx})^n## is kinda like shorthand for "do that differentiation thing ##n## times". So ##\frac{d^ny}{dx^n}=\frac{d^n}{dx^n}[y]## is notation for "do that differentiation thing to ##y## ##n## times" or, in the case of the LHS, "the result of having done that differentiation thing to ##y## ##n## times".
     
  6. Aug 30, 2014 #5
    @gopher_p

    Thanks for your reply! I always knew that Leibniz's notation was somewhat of needed abuse. I think my problem was that my Calculus classes did not cover the Epsilon-Delta definition of limit (and in turn, the limit nature of the derivative) in detail, so I got too used to the beauty the notation can give. I just thought that you could define a second order differential ##d^2y## in the same way you can with ordinary differentials ##df(x)=f'(x)dx##. I guess this is not possible. Your reply made me realize what I was doing wrong. Thanks again!
     
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