Page 1 of 1

9367 44

Posted: Thu Feb 16, 2012 12:47 am
by yoyobarn
Hi,

Could someone help me with 9367 Q44?

I tried using L'Hopital's Rule twice (differentiating with respect to h), but ended up getting 8P''(x).

Thank you very much.

Re: 9367 44

Posted: Thu Feb 16, 2012 2:37 pm
by aaroncraig
You probably just make a careless error:

Image

So,

Image

L'Hopital-itizing again:

Image

Evaluating at h = 0:

Image

So the answer is D, as hoped.

Re: 9367 44

Posted: Thu Feb 16, 2012 10:29 pm
by yoyobarn
Thanks a lot!

Yes, I made a careless error, forgot to multiply by 0 for the term 2P(x).

I have a question, when we differentiate with respect to h, in the end is the P''(x) with respect to h or x?
Thanks in advance.

Re: 9367 44

Posted: Fri Feb 17, 2012 12:39 pm
by aaroncraig
I'm confused as to what you mean. The only derivatives evaluated in the above solution are taken with respect to h. P(x) is a function of x only. P(x) has no h-dependence, so dP/dh = 0, which kills the P(x) term. Also, unless I am misunderstanding your question, there is no point in the above solution at which we are multiplying by zero. All I did was apply L'Hopital's rule twice to obtain a limit without infinity in the denominator. After twice differentiating, the denominator becomes simply '2', meaning that the limit of the whole expression is well-defined (finite), allowing us to evaluate at h = 0.

Re: 9367 44

Posted: Sat Feb 18, 2012 1:55 am
by yoyobarn
Sorry my bad,
I meant "dP/dh = 0, which kills the P(x) term", when I wrote "multiply by 0".

My second question is that I am confused about the prime (') in the question.

For example y' can either mean dy/dx or dy/dt, depending on the context.
Hence, my question is that in Q44, does the P''(x) in the final 5 options refer to d^2/dx^2 P(x) or d^2/dh^2 P(x).

I would think that P''(x) would be "with respect to x", but would that contradict the fact that in my working, P was being differentiated with respect to h.

Thanks for clearing my doubts..