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    (Original post by Noble.)
    Not quite.

    y = x^x

    Taking logs you get

    \log(y) = x\log(x)

    When you differentiate, you need to remember that y is a function of x.
    I edited my above post since I couldn't get the latex down, or is that still wrong? .. Oh I see my mistake I differentiated incorrectly, do you get y' = (ln(x) +1)e^(xln(x)) ?

    And thanks revelry26!
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    (Original post by revelry26)
    How many questions are you done with?

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    Well I've only just started, have done the second, about to do the third. How about you, how far have you gotten?

    Btw are there solutions for these questions?
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    (Original post by TheFuture001)
    Well I've only just started, have done the second, about to do the third. How about you, how far have you gotten?

    Btw are there solutions for these questions?
    14. I'm stuck on the fifteenth but I have have a feeling that it's not that hard and I'm just missing some crucial step.

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    (Original post by TheFuture001)
    I edited my above post since I couldn't get the latex down, or is that still wrong? .. Oh I see my mistake I differentiated incorrectly, do you get y' = (ln(x) +1)e^(xln(x)) ?

    And thanks revelry26!
    Yes, but note that e^{x\ln(x)} = e^{\ln(x^x)} = x^x = y

    The cleanest way is just to note that

    \dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{x}{x} + \ln(x) = 1 + \ln(x)

    So,

    \dfrac{dy}{dx} = y(1+\ln(x)) and y = x^x giving

    \dfrac{dy}{dx} = x^x(1+\ln(x))
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    (Original post by Noble.)
    Yes, but note that e^{x\ln(x)} = e^{\ln(x^x)} = x^x = y

    The cleanest way is just to note that

    \dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{x}{x} + \ln(x) = 1 + \ln(x)

    So,

    \dfrac{dy}{dx} = y(1+\ln(x)) and y = x^x giving

    \dfrac{dy}{dx} = x^x(1+\ln(x))
    Oh yes, I see very nice. Thanks

    @revelry26 I'm going to check that out now.
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    (Original post by TheFuture001)
    Oh yes, I see very nice. Thanks

    @revelry26 I'm going to check that out now.
    Please let me know after you're done with that one I could use a hint :P

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    (Original post by revelry26)
    Please let me know after you're done with that one I could use a hint :P

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    The best hint would be to skip 15, it isn't integrable

    Certainly not using pre-university techniques anyway.
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    (Original post by revelry26)
    Please let me know after you're done with that one I could use a hint :P

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    Aha right back at cha :L

    (Original post by Noble.)
    The best hint would be to skip 15, it isn't integrable

    Certainly not using pre-university techniques anyway.
    Oh really? Well thank God for that I couldn't see a breakthrough anywhere.
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    (Original post by Noble.)
    The best hint would be to skip 15, it isn't integrable

    Certainly not using pre-university techniques anyway.
    Isn't integrable or isn't integrable because I don't know the technique yet? Could you please give me a hint regarding the technique? I tried some substitutions but they all lead to bizarre stuff.

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    (Original post by revelry26)
    Isn't integrable or isn't integrable because I don't know the technique yet? Could you please give me a hint regarding the technique? I tried some substitutions but they all lead to bizarre stuff.

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    It can be done using the power series of \ln(x). it still doesn't integrate to a nice function though.
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    (Original post by Noble.)
    It can be done using the power series of \ln(x). it still doesn't integrate to a nice function though.
    Using \ln(1+x) = \frac{x}{1}-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+... \text{ where }-1<x\le 1 ?

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    (Original post by revelry26)
    Using \ln(1+x) = \frac{x}{1}-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+... \text{ where }-1<x\le 1 ?

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    Yes, pretty much. You'd get an answer in terms of a power series, but it's probably the most straight-forward way to do it.
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    (Original post by Noble.)
    Yes, pretty much. You'd get an answer in terms of a power series, but it's probably the most straight-forward way to do it.
    Okay. Makes a little bit of sense. I need to work on that one a bit more. Thank you so much
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    (Original post by revelry26)
    x
    (Original post by TheFuture001)
    x
    Hi! Mind if I join?
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    (Original post by souktik)
    Hi! Mind if I join?
    Not at all we were talking about the Oxford interview questions. Did you have a look at those?

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    (Original post by revelry26)
    Not at all we were talking about the Oxford interview questions. Did you have a look at those?

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    I'm very bad at these graph sketching questions. Any suggestions for where I can practice such questions from? :P
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    Have y'all heard back from your colleges or are you practicing just in case? Merton said on their Facebook that the letters were being sent out every day by subject, so I hope Mathematics comes soon. I've become so on-edge, it's gotten to the point where I have a mini-heart attack everytime my phone makes the email notification sound, only to read it and see it's some irrelevant message.

    Also where do I find these interview questions that are being discussed?
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    (Original post by revelry26)
    Not at all we were talking about the Oxford interview questions. Did you have a look at those?

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    Yeah, I know. That's what I wanted to join.
    Just taking a look.
    The non-calculus questions seem interesting enough.
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    (Original post by souktik)
    Yeah, I know. That's what I wanted to join.
    Just taking a look.
    The non-calculus questions seem interesting enough.
    The calculus ones are pretty easy so far. The others are interesting. I'm still not convinced with my solution for the fifteenth one. :-/

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    (Original post by dutchmaths)
    Have y'all heard back from your colleges or are you practicing just in case? Merton said on their Facebook that the letters were being sent out every day by subject, so I hope Mathematics comes soon. I've become so on-edge, it's gotten to the point where I have a mini-heart attack everytime my phone makes the email notification sound, only to read it and see it's some irrelevant message.

    Also where do I find these interview questions that are being discussed?
    Many colleges are yet to send out Maths invites, it seems. Go back a couple of pages, I believe Noble posted them as an attached docx.
 
 
 
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