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Chain rule =/

  1. Dec 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi all , again i got stuck onto this iffy bit , question is:
    find dy/dx of the curve and hence its coordinates of the statinonary point.
    y= 4 - 2/(2x-1)^2 -x
    2. Relevant equations
    du/dx x dy/du

    3. The attempt at a solution
    well this is how i started off:
    let u = (2x-1) , y= 4-2(u)^-2 - x
    du/dx= 2;dy/du:4u^-3-1

    [8u^(-3) -2]
    to find stationary points i will let dy/dx= 0 and solve the equation ?am i on the right track so far?
  2. jcsd
  3. Dec 12, 2009 #2
    oo i think i have made a mistake.. i should have multiplied gradient u (2) with the first term for the gradient of dy/du ?
    EDIT:sorry for being such a spammer,i managed to solve this question but grr.. i am stuck on the next one.
    question is:
    curve y = (3x-2)^3 - 5x^2
    find dy/dx and stationary points.

    My attempt :
    dy/dx= 9u^2 - 10x
    we know u= (3x-2), so equation becomes
    9(3x-2)^2- 10x
    81x^2 -108x +26= 0
    divide by 9
    9x^2 -12x -2.8 ( am i over-complicating it?)
    Last edited: Dec 12, 2009
  4. Dec 13, 2009 #3


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    Science Advisor

    Check your arithmetic! 9(9x^2- 12x+ 4)- 10x= 81x^2- 108x+ 36- 10x.

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