The following is supposed to draw a connection between Lp norm and entropy: f is a probability density below:
$$\left. \frac{d}{dp}\right|_{p=1} \left[\int f^p \,d\mu\right]^{1/p} =\int f \log(f) \,d\mu$$
My derivation always reveals the LHS to be 0. What I did wrong?
$$\frac{d}{dp} \left[ \int f^p d\mu\right]^{1/p} = \left(\int f^p \, d\mu\right)^{1/p} \log \left(\int f^p \, d\mu\right) \cdot \left(-p^{-2} \right) \cdot \int f^p \log(f) \, d\mu$$
which is $0$ when $p=1$ (since $\log\left(\int f^p \, d\mu\right)=\log1=0$).
Write $h(p)=\int f^pd\mu$. $h'(p)=\int f^p\log(f)d\mu.$
Then we have $$\frac{d}{dp}(h(p)^{1/p})=\frac{d}{dp}(e^{\frac{\log(h(p))}{p}})=e^{\frac{\log(h(p))}{p}} \frac{\frac{p}{h(p)}h'(p)-\log(h(p))}{p^2}$$
When p=1, the above evaluates to $$h'(1)=\int f^p \log(f)d\mu.$$