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Factorising a differntiated equation

  1. Sep 26, 2007 #1
    Hello there,

    I'm quite weak in factorising functions,especially those with indices.
    I would appreciate any link on these kind of factorisation or even some tutoring on it.

    this is a question I'm stuck in.

    y=(2x+3)(4x-3)^1/2, show dy/dx can be written in the form kx/(4x-3)^1/2

    my solution:

    dy/dx =v(du/dx) + u(dv/dx)

    where u=2x+3
    dv/dx= -2(4x-3)^-1/2

    then, dy/dx=[(4x-3)^1/2 ]2 -(-2)(2x+3)[(4x-3)^-1/2]
    = 2{(4x-3)^1/2 - (2x+3)(4x-3)^-1/2}

    = . . . .

    I am not sure what to do after that
    Last edited: Sep 26, 2007
  2. jcsd
  3. Sep 26, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    There is no question there at all! Do you mean you are asked to find dy/dx in that form?

    No, dv/dx= (1/2)(4x-3)^(-1/2)

    First, as I said, dv/dx= (1/2)(4x-3)^(-1/2), not what you have.

    Making that correction, dy/dx= 2(4x-3)^(1/2)+ (1/2)(2x+3)(4x-3)^(-1/2).

    It might help you to note that (4x-3)^(1/2)= (4x-3)/(4x-3)^(1/2) and, of course, (4x-3)^(-1/2)= 1/(4x-3)^(1/2). In other words, by replacing (4x-3)^(1/2) with 2(4x-3)/2(4x-3)^(1/2) you have "common denominators":

    dy/dx= 4(4x-3)/(2(4x-3)^(1/2))+ (2x+3)/(2(4x-3)^(-1/2)
    can you finish that?
  4. Sep 26, 2007 #3
    I do not understand why is it so

    does not the chain rule state that dy/dx=dy/du *du/dx

    hence, when y= U^n

    dy/dx= (nu^n-1) * (du/dx)


    v = (4x-3)^1/2

    then dv/dx=1/2(4x-3)^(1/2 -1 ) *(4)

    (correction to my previous equation with that was dv/dx =-2(4x-3)^-1/2)

    correct me if I am wrong, pls n ty!

    and the second part.....
    I have never learnt that till now that is my Secondory 5 year!(o-levels like)

    enlighten me!
  5. Sep 26, 2007 #4
    Oh oh,
    i got the last part , now(had to work it out in a piece of paper first...oops)
  6. Sep 26, 2007 #5
    but i still do not get the chain rule.
    my text book says that , by the waY.
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