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Multiplying by dX

  1. Mar 12, 2014 #1
    there was this curve where
    Dy/Dx = 3-x/y+4
    so he cross multiplied the equation
    as (3-x)Dx=(y+4)Dy , then he proceeded to integrate the function .
    i dont know if this is correct or not , i mean Dy/dx is just a notation , how can you treat as if it was just a normal fraction ?
    i thought he treated them as if they were differentials
    then he took the integral of both sides , and he treated Dx as if it was just a notation that means integral with respect to X and so with Dy
  2. jcsd
  3. Mar 12, 2014 #2


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    You are right, this does look like an abuse of the notation, but that's the way it is usually written. You can write it out more formally:
    $$\begin{aligned}\frac{dy}{dx} &= \frac{3-x}{y+4}\\
    (y+4)\frac{dy}{dx} &= 3-x\\
    \int(y+4)\frac{dy}{dx}dx &= \int(3-x)dx\\
    \int(y+4)dy &= \int(3-x)dx
    But life is too short to write out the solution to every problem in all that detail.
  4. Mar 12, 2014 #3
    in 3rd > 4th step
    how did (dy/dx)dx become dy ?
  5. Mar 12, 2014 #4


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    You've ##dx/dx## on the right side. One of the ##dx## goes to the left side and voila
    Last edited: Mar 12, 2014
  6. Mar 12, 2014 #5


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    That's an application of the chain rule. You can also define "differentials":
    f'(x)= dy/dx is NOT a fraction but IS the limit of a fraction:
    [tex]\frac{dy}{dx}= \lim_{h\to 0}\frac{f(x+ h)- f(x)}{h}[/tex]
    So it has properties like a fraction that can be proven by going back before the limit, using the fraction property, then taking the limit.
    We can make use of the fact that the derivative is like a fraction by defining "differentials":
    if y= f(x), so that y'= f'(x), define dy= f'(x)dx where dx and dy are "differentials".

    For example, if y is a function of x and x is a function of t, we can write the "chain rule" as
    [tex]\frac{dy}{dt}= \left(\frac{dy}{dx}\right)\left(\frac{dx}{dt}\right)[/tex]
    That is commonly proved by, as I said, going back before the limits to write
    [tex]\frac{y(x(t)+ h(t))- y(x(t))}{h(t)}\frac{h(t+ j)- h(t)}{j}[/tex]
    and cancelling. But with "differential" notation we can think of it as actually cancelling the "dx" terms.
  7. Mar 12, 2014 #6
    then the dX in the end of every integral is not then just a sign to know what are you integrating with respect to , it is actually a differential .. hmmm.. thanks :D
  8. Mar 12, 2014 #7
    Maybe an infinitesimal quantity would be a better description. A differential would be d/dx.
  9. Mar 12, 2014 #8


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    dy/dx can be considered to be a ratio of two infinitesimal numbers. This is the way calculus was traditionally understood until about 1890, and Abraham Robinson provided some theoretical justification around 1960. There is a nice freshman calc book by Keisler that takes this approach, and it's free online: http://www.math.wisc.edu/~keisler/calc.html
  10. Mar 12, 2014 #9


    Staff: Mentor

    No. d/dx is a derivative operator. It is not a differential.
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