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    (Original post by jayshah31)
    Looks fine. I don't think you really need to simplify it from there.
    Okay the next one is:

     y = \frac{1}{x}(\frac{1}{2} - x)

    I'm just not sure what to do!

    Sorry for all these questions. I realise this is basic stuff, and just regret picking Maths to do this year now!
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    (Original post by Darkest Knight)
    Hi -

    I've got an exercise to do on this, with 3 sets of different types of questions, and I really don't understand it at all!

    The first type goes like this:

    Differentiate each expression with respect to the given variable

    a. y = x(x+1)
    b. s = t^2(1-t)
    j. y = (x^3 - 1)(x - 1)

    I really have no idea how to do them. I thought I knew how to differentiate, well I know I'm meant to multiply the "other" number by the power, and then - 1 from the original power?

    Hope you can help
    try expanding the brackets?

    i think you can also use product rule here, but that's not in C1, so expanding the brackets is the plan :yy:
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    (Original post by Darkest Knight)
    Okay the next one is:

     y = \frac{1}{x}(\frac{1}{2} - x)

    I'm just not sure what to do!

    Sorry for all these questions. I realise this is basic stuff, and just regret picking Maths to do this year now!
    You can use the product rule, but as above, forget about it - it's not C1.

    Use this rule of thumb: Always expand any brackets you have first

    Spoiler:
    Show

     y = \frac{1}{x}(\frac{1}{2} - x)

     y = \frac{1}{2x} - \frac{x}{x}

    As you can see anything divided by itself is 1 so....
     y = \frac{1}{2x} - 1

    Can you take it from there?
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    (Original post by jayshah31)
    You can use the product rule, but as above, forget about it - it's not C1.

    Use this rule of thumb: Always expand any brackets you have first

    Spoiler:
    Show

     y = \frac{1}{x}(\frac{1}{2} - x)

     y = \frac{1}{2x} - \frac{x}{x}

    As you can see anything divided by itself is 1 so....
     y = \frac{1}{2x} - 1

    Can you take it from there?
    I wrote it out wrong lol.

    It's actually:


     y = \frac{1}{x}(\frac{1}{x} - x)

    Sorry, does that just simplify to   y = \frac{1}{x^2} - \frac{x}{x^2}
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    (Original post by Darkest Knight)
    I wrote it out wrong lol.

    It's actually:


     y = \frac{1}{x}(\frac{1}{x} - x)

    Sorry, does that just simplify to   y = \frac{1}{x^2} - \frac{x}{x^2}
    Bingo. I had a hunch you did, as I would really only be able to differentiate what I came up with in C3

    Yes, your simplification is correct, remember that \frac{x}{x^2} will cancel to \frac{1}{x}. If you can't immediately differentiate it, I don't blame you. There's a quick nudge you may need ...
    Spoiler:
    Show
    \frac{1}{x^2} = x^{-2}

    :rolleyes:


    Jay
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    (Original post by jayshah31)
    Bingo. I had a hunch you did, as I would really only be able to differentiate what I came up with in C3

    Yes, your simplification is correct, remember that \frac{x}{x^2} will cancel to \frac{1}{x}. If you can't immediately differentiate it, I don't blame you. There's a quick nudge you may need ...
    Spoiler:
    Show
    \frac{1}{x^2} = x^{-2}

    :rolleyes:


    Jay
    Thanks. So I differentiate them, and get  -2x^-^3 + x^-^2 Is that right up to there? I realise I need to simplify it, the books answer is  - \frac{2}{x^3} and I don't know how to get there!
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    (Original post by jayshah31)
    Bingo. I had a hunch you did, as I would really only be able to differentiate what I came up with in C3

    Yes, your simplification is correct, remember that \frac{x}{x^2} will cancel to \frac{1}{x}. If you can't immediately differentiate it, I don't blame you. There's a quick nudge you may need ...
    Spoiler:
    Show
    \frac{1}{x^2} = x^{-2}

    :rolleyes:


    Jay
    Sorry, no, I incorrectly said yes the first time. When you expand, your last term is -\frac{x}{x} which is -1.
    Spoiler:
    Show

     -2x^-^3

    = -2 \times x^{-3}

    = -2 \times \frac{1}{x^{3}}

    = \frac{-2}{x^3}


    Jay
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    (Original post by jayshah31)
    Sorry, no, I incorrectly said yes the first time. When you expand, your last term is -\frac{x}{x} which is -1.
    Spoiler:
    Show

     -2x^-^3

    = -2 \times x^{-3}

    = -2 \times \frac{1}{x^{3}}

    = \frac{-2}{x^3}


    Jay
    Okay, that makes sense.

    Okay onto the next question. Could you check this for me please: does (1-t)(1-2t)(1-3t) =  -6t^2 - 9t^3 - 3t ? Not really sure of the "correct" method to expand 3 sets of brackets, I normally use FOIL, for two sets.. Thanks
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    I think I got something different. Expand two brackets, then multiply that by the 3rd.

    (1-t)(1-2t)(1-3t)\\

(1-3t+2t^2)(1-3t)\\

1-3t-3t+9t^2+2t^2-6t^3
    Either that, or spot the mistake I made above (you can simplify mine, I just can't be bothered with the latex)
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    (Original post by Dom K. Jordan)
    It is the same.
    I know it is, now anyway. I made that post before you made yours (well, Latex took up a lot of time).
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    (Original post by Dom K. Jordan)
    I would just multiply out two brackets first then multiply the third by the resultant quadratic.

    (1-t)(1-2t) = (1 - 3t + 2t^2) \times (1-3t) = 1 - 6t + 11t^2 - 6t^3
    Dom,

    Please be careful when using = signs.

    1 + 2 = 3 \times 5 = 15

    This sort of thing makes teachers want to hang themselves.

    Mr M
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    (Original post by Dom K. Jordan)
    Sorry my mistake I thought you were talking to me at first, but I changed my post as well when I realised you weren't.
    No problem. The = signs are a good point. I learned the not-so-easy way
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    (Original post by jayshah31)
    I think I got something different. Expand two brackets, then multiply that by the 3rd.

    (1-t)(1-2t)(1-3t)\\

(1-3t+2t^2)(1-3t)\\

1-3t-3t+9t^2+2t^2-6t^3
    Either that, or spot the mistake I made above (you can simplify mine, I just can't be bothered with the latex)
    Sorry, this will sound stupid. Just not sure, how to go about multiplying the first set of brackets by second? Normally I'd use the "FOIL" method, but this won't work, as there are 3 terms in the first bracket, and 2 in the second.

    THanks
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    First, Outside, Inside, Last. Method for expanding two binomials that are multiplied together.
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    (x-3)(2x+7)

    First: x*2x
    Outside: 7*x
    Inside: -3*2x
    Last: -3*7

    (2x^2+x-21)

    Mr M is just out to sabotage my efforts :ninja:
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    (Original post by D-Day)
    (x-3)(2x+7)

    First: x*2x
    Outside: 7*x
    Inside: -3*2x
    Last: -3*7

    (2x^2+4x-21)
    You might want to collect those like terms again DDay!
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    (Original post by Mr M)
    You might want to collect those like terms again DDay!
    ahaha. I missed the 2 coefficient :ninja:
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    (Original post by Dom K. Jordan)
    (a + b + c)(d + e) = ad + ae + bd + be + cd + ce
    Literally the same as if there's two. What's FOIL?

    So starting from the beginning
    (a + b)(c + d)(e + f) = (ac + ad + bc + bd)(e + f) = ace + acf + ade + adf + bce + bcf + bde + bdf
    And collect like terms
    Thanks, done that and I get: 1 - 9t + 5t^2?

    Thing is, I just won't remember how to do that in an exam!
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    (Original post by Dom K. Jordan)
    Its the same, i don't get why this FOIL method is used it just makes it overcomplicated. I don't like those kind of teaching methods because you get stuck when you go into more advanced questions.

    Remember all your doing is multiply each term from one bracket with each in the other, if you remember it that way you have a general rule for any number of brackets.
    Thanks is that answer I got right?

    I differentiated it and got 5t^2 + 10t, but the book says the differentiated answer is: -18t^2 +22 -6
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    (Original post by Darkest Knight)
    Thing is, I just won't remember how to do that in an exam!
    If you are not confident with expanding brackets, you could use a grid to organise your work.

    This one shows the expansion of (x-3)(2x+7)

    Name:  Grid multiplication.JPG
Views: 83
Size:  21.5 KB

    You still need to collect like terms at the end.
 
 
 
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