1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook