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Numbers of Arrangements

https://isaacphysics.org/questions/differentiation_6_5?board=7e6911fd-27e1-497f-8d06-b00d3a634bc8&stage=a_level

For part b of this question, once you get an expression for dln(g)/dm1, how do you then get an expression for dg/dm1

For reference my expression for dln(g)/dm1:
ln [ (n1+m1)(m-m1)/m1(n-n1+m-m1) ] -> m1=n1*m/n

so for dg/dm1, do you simply remove the ln(...) or..?
Im not sure why you would be able to do that or if there is another method.
Any help greatly appreciated as always.
Reply 1
Original post by mosaurlodon
https://isaacphysics.org/questions/differentiation_6_5?board=7e6911fd-27e1-497f-8d06-b00d3a634bc8&stage=a_level
For part b of this question, once you get an expression for dln(g)/dm1, how do you then get an expression for dg/dm1
For reference my expression for dln(g)/dm1:
ln [ (n1+m1)(m-m1)/m1(n-n1+m-m1) ] -> m1=n1*m/n
so for dg/dm1, do you simply remove the ln(...) or..?
Im not sure why you would be able to do that or if there is another method.
Any help greatly appreciated as always.

Its just the usual log derivative / chain rule so
d ln(g(m1)) / dm1 = 1/g dg/dm1
Oh so
d^2 lng / dm1^2 = (d lng/dg)^2 x d^2g/dm1^2?
or would you use chain rule for second derivative?
Reply 3
Original post by mosaurlodon
Oh so
d^2 lng / dm1^2 = (d lng/dg)^2 x d^2g/dm1^2?
or would you use chain rule for second derivative?

Ive not worked it through, but the question reads (the "hence") like they want you to differentiate dg/dm1 to get the second derivative directly, rather than using the dln(g)/dm1. Though if that is troublesome, you could possibly set
h(m1) = dg/dm1 = expression
and take logs, differentiate, then use the chain rule to get dh/dm1 which is the desired second derivative of g.
(edited 1 month ago)
Im not going to lie it took me a long time to solve the question with all those terms (and a lot of double checking each step with wolframalpha) but it finally worked - thank you very much
Reply 5
Original post by mosaurlodon
Im not going to lie it took me a long time to solve the question with all those terms (and a lot of double checking each step with wolframalpha) but it finally worked - thank you very much

No problem. Not worked it through but it looks like a bit of a slog. The useful trick is that you can use ln() to somewhat simplify differentiating a fraction/product of terms and of course stirlings formula. When you had to show that at a stationary point g_m X/(...), you only really need half the second derivative formula as its g*something and using the product rule, the first term g'*something will be zero.

In a very simple case you could do
y = f/g
ln(y) = ln(f) - ln(g)
1/y y'= 1/f f' - 1/g g'
though its a neat way to write the quotient rule, multiply through by y to get the usual form so
y' = f'/g - fg'/g^2 = ...
So one of several ways to get the quotient rule.
Could I just double check my reasoning?
dg/dm1 = g * dlng/dm1
d^2g/dm1^2 = dg/dm1 * dlng/dm1 + g * d^2lng/dm1^2
I just assumed the first term would go to 0, but didnt really check with why - do you mind please explaining?

Also thank you for your quotient rule derivation - are you saying that would be quicker for differentiating dln(g)/dm1 -> d^2ln(g)/dm1^2 than just normally differentiating it (probably because that took quite a while) - if so what would you use as f and g?
Reply 7
Original post by mosaurlodon
Could I just double check my reasoning?
dg/dm1 = g * dlng/dm1
d^2g/dm1^2 = dg/dm1 * dlng/dm1 + g * d^2lng/dm1^2
I just assumed the first term would go to 0, but didnt really check with why - do you mind please explaining?
Also thank you for your quotient rule derivation - are you saying that would be quicker for differentiating dln(g)/dm1 -> d^2ln(g)/dm1^2 than just normally differentiating it (probably because that took quite a while) - if so what would you use as f and g?

The first term dg/dm1*dlng/dm1 is zero at a stationary point as dg/dm1 is zero, but in general its obviously nonzero. For differentiating dln(g)/dm1 to get the second derivative, I would have thought that was what the question was hinting at as you want to show that at a stationary point the second derivative is 0 + g*#. Not sure what # or how easy it is to differentiate is as Ive not worked it through, but Id have thought that would be the way to go.

The usual quotient rule was just a simple example to illustrate that taking logs and then differentiating is sometimes a useful trick.
Oh right, that is actually crucial for this problem, I just completely missed it and got lucky.
Thank you for the trick with the quotient rule, Ill try to keep that in mind for future problems.

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