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    (Original post by Zacken)
    Practice.
    (Original post by RDKGames)
    Practice a lot.
    (Original post by ValerieKR)
    I have perfect recall.
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    Practice.
    Lol great, thanks 😂😂😂

    So shouldn't I worry about flash cards and all that stuff?
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    (Original post by asinghj)
    Lol great, thanks 😂😂😂

    So shouldn't I worry about flash cards and all that stuff?
    Definitely don't bother with flash cards. That's the definition of anti-revision for maths.
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    (Original post by RDKGames)
    Definitely don't bother with flash cards. That's the definition of anti-revision for maths.
    👍
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    I don't want annihilation to take place.
    Ba dum tss
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    (Original post by ValerieKR)
    You're not applying to Oxford, are you? -.-
    I am applying to Oxford...
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    Why is the answer +- for this C3 question

    ln x^2 = 8
    2lnx = 8
    lnx = 4
    x= e^4

    But it should be x= +- e^4
    Why is it +- (im not taking the sq rt of 8?)
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    (Original post by kiiten)
    Why is the answer +- for this C3 question

    ln x^2 = 8
    2lnx = 8
    lnx = 4
    x= e^4

    But it should be x= +- e^4
    Why is it +- (im not taking the sq rt of 8?)
    If you exponentiate both sides to begin with you get  x^2=e^8 .
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    (Original post by kiiten)
    Why is the answer +- for this C3 question

    ln x^2 = 8
    2lnx = 8
    lnx = 4
    x= e^4

    But it should be x= +- e^4
    Why is it +- (im not taking the sq rt of 8?)
    You're not getting a negative because of your second step where ln(x) cannot take upon negative x values whereas in the first one the x in ln(x2) can indeed take upon both negative and positive ones. So technically speaking you can go ahead with your method as long as you remember to include the opposite sign of your answer.
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    (Original post by kiiten)
    Why is the answer +- for this C3 question

    ln x^2 = 8
    2lnx = 8
    lnx = 4
    x= e^4

    But it should be x= +- e^4
    Why is it +- (im not taking the sq rt of 8?)
    ln|x|=4 so |x|=e^4
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    (Original post by SeanFM)
    Handy tip, at the end of a pdf link if you type '#page6' for example, it will take the reader to page 6 immediately.
    OK do u know explanation for that answer
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    (Original post by youreanutter)
    OK do u know explanation for that answer
    Please link me to the correct page (that is the reason I told you how to do it), I do not want to filter through a whole PDF to look for a particular reference.

    Thank you
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    So i have this differentiation question,
    Show that  \frac{d}{dx} ((4x-1)\sqrt{4x-1}) = 6\sqrt{4x-1}
    I tried to use product rule, but the u and v are the same and i dont know how to go about it,
    I tried the chain rule, but still have no clue, as there are x terms on both of them.
    So basically i dont know where to get started
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    (Original post by asinghj)
    So i have this differentiation question,
    Show that  \frac{d}{dx} ((4x-1)\sqrt{4x-1}) = 6\sqrt{4x-1}
    I tried to use product rule, but the u and v are the same and i dont know how to go about it,
    I tried the chain rule, but still have no clue, as there are x terms on both of them.
    So basically i dont know where to get started
    product rule
    u=sqrt(4x-1)
    v=4x-1
    derivative of uv = u*v' + u'*v


    alternatively you're finding the derivative of (4x-1)^3/2, chain rule with 4x-1=u
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    (Original post by ValerieKR)
    product rule
    u=sqrt(4x-1)
    v=4x-1
    derivative of uv = u*v' + u'*v


    alternatively you're finding the derivative of (4x-1)^3/2, chain rule with 4x-1=u
    OMG, thank you. I spent the whole day trying to figure this out and was overcomplicating so much, when the answer was so simple😂😂

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    in my defence it wasone of the first differential questions i did after just learning the rules
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    can someone explain how to do 1d?
    http://www.undergraduate.study.cam.a...tion_paper.pdf


    For part C i resolved the forces to make acceleration the subject and integrated it to get velocity, and I got D as the answer for that, not sure how to work out part D, do I have to resolve again and include kv as drag?

    or do I do kv=ma, use the acceleration from previous section?
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    (Original post by B_9710)
    If you exponentiate both sides to begin with you get  x^2=e^8 .
    the question is ln x^2 = 8
    im confused how you got from that to x^2 = e^8

    (Original post by RDKGames)
    You're not getting a negative because of your second step where ln(x) cannot take upon negative x values whereas in the first one the x in ln(x2) can indeed take upon both negative and positive ones. So technically speaking you can go ahead with your method as long as you remember to include the opposite sign of your answer.
    Im confused
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    Hi gonna jump on this thread even though it's a bit late
    I would like to be a helper and a learner lol as I have done C3, C4 and M2 and am now on a gap yahh doing FP1, FP2, S1 and S2. Did pretty good in both C3 and C4 so should be able to be quite helpful there, though maybe not as helpful as people who are really good at maths (i.e. those doing it at university)
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    (Original post by kiiten)
    the question is ln x^2 = 8
    im confused how you got from that to x^2 = e^8



    Im confused
    ln is log in base e, so you can do e to the power of both sides to eliminate the ln, understand?
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    (Original post by metrize)
    can someone explain how to do 1d?
    http://www.undergraduate.study.cam.a...tion_paper.pdf


    For part C i resolved the forces to make acceleration the subject and integrated it to get velocity, and I got D as the answer for that, not sure how to work out part D, do I have to resolve again and include kv as drag?

    or do I do kv=ma, use the acceleration from previous section?
    now I am a future biologist, so maybe not a leading authority on engineering - but I do think I've worked it out.
    It makes sense the block will reach terminal velocity, right, when the forces acting forward = those acting backward.
    So you set up the diagram, mg sin(a) down the slope and uN + kV up the slope, as we are adding in the air resistance term (kV). N = mg cos (a) , so you end up with this;
    mg sin(a) = u mg cos(a) + kV
    and you rearrange it get V out, because that V is gonna be the terminal velocity. At lesser values of V, the forces down slope would be greater and the object would accelerate. Do you agree with that?? Hope you can understand it lol (I had to use a for alpha and u for mu)
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    (Original post by kiiten)
    the question is ln x^2 = 8
    im confused how you got from that to x^2 = e^8
    That is worrying if you're confused on that. Remember that \ln(x^2)=\log_e(x^2), and you SHOULD be able to know what happens here after Core 2.

    (Original post by kiiten)
    Im confused
    Not sure how else to explain it to you. You should know that you cannot have a logarithm of a negative number, therefore \ln(x) cannot have x as negatives. When it comes to \ln(x^2) however, then you CAN have negative x values because negatives/positives squared just become positive.

    When you go from \ln(x^2) to 2\ln(x) you are essentially getting rid off all the negative x-values hence getting rid off one of the solutions. Get it?
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    (Original post by k.russell)
    ln is log in base e, so you can do e to the power of both sides to eliminate the ln, understand?
    Ohhhh so you do e^(ln x^2) = e^8
    X^2 = e^8
    X = sqrt e^8

    ?? Wait ive gone wrong, sorry my minds just gone blank
 
 
 
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