For the case of the Debye's model of a solid, the free energy is defined as
I don't know how to calculate the derivative above. I have got just the answer, but intermediate calculus don't. Another think, in my exercise is specified to use the Leibniz role about derivative of a integral that have a function depends of two variable. Please help.
Please note the Debye functions $D_n (z)$ are given by $$D_n (z) = \frac{n}{z^n} \int^{z}_0 \frac{x^n}{e^x - 1} \, dx.$$ So when $n = 3$ we have $$D_3 (z) = D(z) = \frac{3}{z^3} \int^z_0 \frac{x^3}{e^x - 1} \, dx.$$
Given $$F(T) = F_0 + N \nu k_B T \left [3 \ln (1 - e^{-T_D/T}) - D(T_D/T) \right ],$$ to find $$S = - \frac{\partial F}{\partial T},$$ note that $$\frac{\partial}{\partial T} \left [3 \ln (1 - e^{-T_D/T}) \right ] = - \frac{3 T_D}{T^2 (e^{T_D/T} - 1)},$$ and \begin{align*} \frac{\partial}{\partial T} D \left (\frac{T_D}{T} \right ) &= \frac{\partial}{\partial T} \left \{\frac{3T^3}{T^3_D} \int^{T_D/T}_0 \frac{x^3}{e^x - 1} \, dx \right \}\\ &= \frac{9 T^2}{T^3_D} \int^{T_D/T}_0 \frac{x^3}{e^x - 1} \, dx + \frac{3 T^3}{T^3_D} \frac{\partial}{\partial T} \int^{T_D/T}_0 \frac{x^3}{e^x - 1} \, dx. \end{align*} where the product rule has been used. The derivative of the integral to the right can be found using the Leibniz rule for differentiating under the integral sign, namely $$\frac{d}{dT} \int^{b(T)}_{a(T)} f(T, x) dx = f(T,b(T)) \cdot \frac{d}{dT} b(T) - f(T, a(T)) \cdot \frac{d}{dT} a(T) + \int^{b(T)}_{a(T)} \frac{\partial}{\partial T} f(T,x) dx.$$ In our particular case, as $a(T) = 0$ and $f(T,x) = f(x)$ only, Leibniz's rule reduces to $$\frac{d}{dT} \int^{b(T)}_0 f(x) \, dx = f(b(T)) \cdot \frac{d}{dT} b(T).$$ As $b(T) = T_D/T$, $b'(T) = -T_D/T^2$, and we have $$\frac{\partial}{\partial T} \int^{T_D/T}_0 \frac{x^3}{e^x - 1} \, dx = - \frac{T^4_D}{T^5 ( e^{T_D/T} - 1)}.$$ Thus \begin{align*} \frac{\partial}{\partial T} D \left (\frac{T_D}{T} \right ) &= \frac{3}{T} \cdot \frac{3}{(T_D/T)^3} \int^{T_D/T}_0 \frac{x^3}{e^x - 1} \, dx - \frac{3 T_D}{T^2 ( e^{T_D/T} - 1)}\\ &= \frac{3}{T} D \left (\frac{T_D}{T} \right ) - \frac{3 T_D}{T^2 (e^{T_D/T} - 1)}. \end{align*}
Now returning to the expression for $F$, on differentiating partially with respect to $T$ we have \begin{align*} \frac{\partial F}{\partial T} &= N \nu k_B \left [ 3 \ln (1 - e^{-T_D/T}) - D(T_D/T) \right ] \\ &\quad + N \nu k_B T \left [-\frac{3 T_D}{T^2 (e^{T_D/T} - 1)} - \frac{3}{T} D \left (\frac{T_D}{T} \right ) + \frac{3 T_D}{T^2 (e^{T_D/T} - 1)}\right ]\\ &= N \nu k_B \left [3 \ln (1 - e^{-T_D/T}) - 4 D (T_D/T) \right ], \end{align*} or $$S = N \nu k_B \left [-3 \ln (1 - e^{-T_D/T}) + 4 D (T_D/T) \right ],$$ as required to show.