# Derivation of Wien's displacement law

I know this might be a silly question, but is it necessary to know Planck's Law in order to show that $\lambda_{max}\propto\frac{1}{T}$? If you set $$u(\lambda,T)=\frac{f(\lambda T)}{\lambda^5}$$ then $$\frac{\partial u}{\partial \lambda} = \frac{1}{\lambda^5}\frac{\partial f}{\partial \lambda}-\frac{5}{\lambda ^4}f=0$$

$$\frac{\frac{\partial f}{\partial \lambda}}{f} = 5\lambda$$

But I am stuck here because if I integrate the L.H.S. $$\int \frac{\frac{\partial f}{\partial \lambda}}{f} d\lambda = \log (f(\lambda T)) = \frac{5}{2}\lambda^2$$

• The notation $\partial f/\partial \lambda$ does not make sense if $f=f(\lambda T)$ is only a function of the product. This should be notated as $f'(\lambda T)$, or (if you absolutely must have partial-derivative notation) as $\frac{\partial f}{\partial (\lambda T)}$. – Emilio Pisanty Feb 2 '18 at 11:01

$$\frac{\partial u}{\partial \lambda} = \frac{\color{red} T}{\lambda^5}\frac{\partial f}{\partial \lambda}-\frac{5}{\lambda^{\color{red} 6}}f=0$$
• So $5\frac{f}{\frac{\partial f}{\partial \lambda}}$ must be a constant. However I don't see why this should be true for any $f$ – math4everyone Feb 3 '18 at 4:44