Partial derivatives help

Question

I'm having problems finding the following:

say F = -kTN*ln(2*cosh(βμB))

where everything is a constant and β=(1/kT)

I want to find S = -(∂F/∂T) and the answer should be S = kN[ln(2*cosh(βμB)) - βμB*tanh(βμB)] which I don't get

Then using that equation of S I want to find C = T* (∂S/∂T) which I also don't know how to do. There's a hint to use the chain rule applied to β=(1/kT)

Please help if you can! Thank you

Answer 1

I'm going to do half of it, just finding S.  Finding C is very similar, more of the same messy calculations.  You just have to be careful not to miss anything when using the chain rule, and when combining constants.

 

First let's substitute in for β so we have the complete equation.

 

F = -kTN * ln (2*cosh (βμB)) = -kTN * ln (2 cosh (μB/kT))

 

Finding ∂F/∂T will use the product rule, because there are two terms with T that are multiplied by each other, and the chain rule, because one of the terms is a compound function of T.  So...

 

Start with the product rule...

∂F/∂T = -kTN * ∂/∂T (ln (2 cosh (μB/kT))) + ln (2 cosh (μB/kT)) ∂/∂T (-kTN)

 

For the first term, you use the chain rule several times to get the partial of the ln expression.  The second term is easier...

= -kTN * (1 / (2 cosh (μB/kT))) * 2 sinh (μB/kT) * (-1)μB/(kT)² * k + ln (2 cosh (μB/kT)) * (-kN)

 

Rearrange terms.  In the first term, move all the constants together and move sinh in front of cosh.  In the second term, just move the constant to the front. 

= (-μB/k²T²) * k * (-kTN) * 2 sinh (μB/kT) / (2 cosh (μB/kT))  - kN ln (2 cosh (μB/kT))

 

Cancel out and combine the constants in the first term.  But leave a kN out by itself.  Also remember that sinh/cosh = tanh.

= (μB/kT) * (kN) * tanh (μB/kT) - kN ln (2 cosh (μB/kT))

 

Factor out kN from both terms...

= kN [μB/kT tanh (μB/kT) - ln (2 cosh (μB/kT))]

 

Put β back in for 1/kT...

= kN [βμB tanh (βμB) - ln (2 cosh (βμB))]

 

Since we started out finding ∂F/∂T, we have to negate it to get S:

 

S = -∂F/∂T = kN [ln (2 cosh (βμB)) - βμB tanh (βμB)]

 

Finding ∂S/∂T then goes over a lot of the same ground, because ln (2 cosh (βμB)) shows up again.  But there is some new work to do to find ∂/∂T [βμB tanh (βμB)]