Oxford MAT 2013/2014

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 $x$.

How many questions are you done with?

Posted from TSR Mobile

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?

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

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

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

Posted from TSR Mobile

Please let me know after you're done with that one I could use a hint :P

Posted from TSR Mobile

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

Certainly not using pre-university techniques anyway.

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.

Posted from TSR Mobile

It can be done using the power series of $\ln(x)$. it still doesn't integrate to a nice function though.

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

Posted from TSR Mobile

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.

x

x

Hi! Mind if I join?

Not at all we were talking about the Oxford interview questions. Did you have a look at those?

Posted from TSR Mobile

Not at all we were talking about the Oxford interview questions. Did you have a look at those?

Posted from TSR Mobile

Yeah, I know. That's what I wanted to join.
Just taking a look.
The non-calculus questions seem interesting enough.

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?