Show that $\frac {\partial B} {\partial T} =$ $\frac{c}{(\exp\frac{hf}{kT}-1)^2}\frac{hf}{kT^2}$

Question

Find an expression for $\frac {\partial B} {\partial T}$ applied to the Black-Body radiation law by Planck: $$B(f,T)=\frac{2hf^3}{c^2\left(\exp\frac{hf}{kT}-1\right)}$$

The correct answer is $\frac {\partial B} {\partial T} =$ $\frac{c}{(\exp\frac{hf}{kT}-1)^2}\frac{hf}{kT^2}$

Could anyone please show me the steps involved to obtain this answer?

Answer 1

$$B(f,T)=\frac{2hf^3}{c^2}\frac{1}{e^\frac{hf}{kT}-1}$$

$$\frac{\partial B}{\partial T}=\frac{2hf^3}{c^2}\frac{\partial}{\partial T}\left(\frac{1}{e^\frac{hf}{kT}-1}\right)=$$

$$=-\frac{2hf^3}{c^2}\left(\frac{1}{e^\frac{hf}{kT}-1}\right)^2\frac{\partial}{\partial T}\left(e^\frac{hf}{kT}-1\right)=$$

$$=-\frac{2hf^3}{c^2}\left(\frac{1}{e^\frac{hf}{kT}-1}\right)^2e^\frac{hf}{kT}\frac{\partial}{\partial T}\left(\frac{hf}{kT}\right)=$$

$$=\frac{2h^2f^4}{kc^2}\frac{1}{T^2}\frac{e^\frac{hf}{kT}}{\left(e^\frac{hf}{kT}-1\right)^2}$$

Are you sure that the correct answer is actually correct?

Answer 2

My suggestion is to look at $\log B$. Take the derivate of the logged function wrt to $T$ and multiply by $B(T)$.