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2000 Differentiation?

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  • 02-06-2008 8:13pm
    #1
    Fuascailt
    Closed Accounts Posts: 268 ✭✭


    If anyones, got a chance, the last c part of question 6 is killing me, and its not on e-xamit.ie or you cant get the marking scheme. I got it kept getting X1=X2, but its not, and then got X1+X2=0... Thanks :D


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    #2
    turgon
    Closed Accounts Posts: 3,762 ✭✭✭turgon


    Honours or pass??


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    #3
    Cokehead Mother
    Closed Accounts Posts: 773 ✭✭✭Cokehead Mother


    It's honours. OP could you post your workings?


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    #4
    turgon
    Closed Accounts Posts: 3,762 ✭✭✭turgon


    If it is honours...

    Get the two asymptotes, and find the point of intersection. I think it is (-1,1).

    Then get any point (x1,f(x1)) (which is on the curve) and use central symmetry through the point of intersection.
    Let the image point = (x2,f(x2)). You should then be able to manipulate it to get x1 + x2 + 2 = 0.


    This works on the principal that the graph is symmetrical. Tough question, and it came up in the exam papers before. As my app maths teacher would say "look at that long term"!


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    #5
    Timans
    Registered Users Posts: 2,370 ✭✭✭Timans


    If dy/dx = -1/(x+1)^2

    Then sub x1 and x2 in for x and let them equal each other as they're parallel.

    -1/(x1+1)^2 = -1/(x2+1)^2

    (x2+1)^2 = (x1+1)^2

    x2+1 = x1+1 or x2+1 = -1(x1+1)
    x2 = x1 or x1+x2+2=0

    As x2 is not = x1. ANS = x1 + x2 +2 = 0


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    #6
    Yuugib
    Registered Users Posts: 432 ✭✭Yuugib


    Em.. another qs on this topic..

    how would you differentiate the following?

    x^x ? found that in my notes.. but there is no solution :( .. i m confused again :(


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    #7
    cartman444
    Closed Accounts Posts: 125 ✭✭cartman444


    i think it X to the power of 2x-2


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    #8
    Cokehead Mother
    Closed Accounts Posts: 773 ✭✭✭Cokehead Mother


    Yuugib wrote: »
    Em.. another qs on this topic..

    how would you differentiate the following?

    x^x ? found that in my notes.. but there is no solution :( .. i m confused again :(

    logarithmic differentiation

    y = x^x

    lny = lnx^x

    lny = xlnx

    1/y (dy/dx) = lnx + x/x

    dy/dx = y(lnx + 1)

    dy/dx = x^x(lnx + 1)


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    #9
    Yuugib
    Registered Users Posts: 432 ✭✭Yuugib


    aw yeeessss :) thank you :) .. ah.. hate those "blonde moments" :D


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    #10
    Timans
    Registered Users Posts: 2,370 ✭✭✭Timans


    Cokehead Mother wrote: »
    logarithmic differentiation

    y = x^x

    lny = lnx^x

    lny = xlnx

    1/y (dy/dx) = lnx + x/x

    dy/dx = y(lnx + 1)

    dy/dx = x^x(lnx + 1)
    God almighty.

    All these things are popping up on boards that I have actually never seen before. Surely, that won't come up on the LC?


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    #11
    Fuascailt
    Closed Accounts Posts: 268 ✭✭Fuascailt


    Thanks guys


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    #12
    eoins2345
    Closed Accounts Posts: 154 ✭✭eoins2345


    http://studentxpress.ie/papers/line.htm


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