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

Question

Find an expression for $\frac {\partial B} {\partial T}$ applied to the Black-Body radiation law by Planck:

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

The correct answer (I believe) is $$\frac{\partial B}{\partial T}= \frac{2h^2f^4}{kc^2}\frac{1}{T^2}\frac{e^\frac{hf}{kT}}{\left(e^\frac{hf}{kT}-1\right)^2}$$ as given by a member of this site in an earlier thread.

The problem is that $\frac {\partial B} {\partial T} =$ $\frac{c}{(e^\frac{hf}{kT}-1)^2}\frac{hf}{kT^2}e^\frac{hf}{kT}$ as given in my text.

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

Which expression (if any) for $\frac {\partial B} {\partial T}$ is correct?

Thanks

Answer 1

Make your life easier writing$$B=\frac{2hf^3}{c^2\left(e^\frac{hf}{kT}-1\right)}=\frac{\alpha}{e^{\beta}-1}$$ using $\alpha=\frac{2hf^3}{c^2}$ and $\beta=\frac{hf}{kT}$.

So $$\frac{dB}{dT}=\frac{dB}{d\beta}\times\frac{d\beta}{dT}$$ Now $$\frac{dB}{d\beta}=-\frac{\alpha e^{\beta }}{\left(e^{\beta }-1\right)^2}$$ $$\frac{d\beta}{dT}=-\frac{hf}{kT^2}$$ So $$\frac{dB}{dT}=\frac{hf}{kT^2}\frac{\alpha e^{\beta }}{\left(e^{\beta }-1\right)^2}$$ and $$\alpha\frac{hf}{kT^2}=\frac{2hf^3}{c^2}\frac{hf}{kT^2}=\frac{2h^2f^4}{c^2k T^2}$$ and $$\frac{dB}{dT}=\frac{2h^2f^4}{c^2k T^2}\frac{e^{\frac{hf}{kT} }}{\left(e^{\frac{hf}{kT} }-1\right)^2}$$ which is the correct answer.