I have the following functional
$F(E(k,t))=\int_0^{k}E(\mu,t)\mu^2d\mu\int_k^{\infty}E(p,t)^{1/2}p^{-3/2} dp $
I would like to compute $F'(E_0(k,t))$, the Fréchet derivative of $F$ at $E_0(k,t)$.
If I call
$A(E(k,t))=A(k,t)=\int_0^{k}E(\mu,t)\mu^2d\mu $
$B(E(k,t))=B(k,t)=\int_k^{\infty}E(p,t)^{1/2}p^{-3/2}dp $
Using the product rule,I know
$F'(E_0)=A'(E_0)B(E)+A(E)B'(E_0)$
Now using the method proposed in this answer.
$A'=\frac{d}{ds}A(E_0+sE_1)|_{s=0}=\frac{d}{ds}\int_0^{k}(E_0+sE_1)\mu^2d\mu |_{s=0}=\int_0^{k}(E_1)\mu^2d\mu$
$B'=\frac{d}{ds}B(E_0+sE_1)|_{s=0}=\int_k^{\infty}(E_0+sE_1)^{1/2}p^{-3/2}dp |_{s=0}=\int_k^{\infty}\frac{E_1}{2}(E_0)^{-1/2}p^{-3/2}dp$
so my result is
$F'(E_0)=B(E)\int_0^{k}(E_1)\mu^2d\mu+A(E)\int_k^{\infty}\frac{E_1}{2}(E_0)^{-1/2}p^{-3/2}dp$
I know that this result is wrong because the good one is
$F'(E_0)=B(E)\int_0^{k}\frac{E_1(\mu,t)}{E_0(\mu,t)}\dot{A}(\mu,t)d\mu-\frac{1}{2}A(k,t)\int_k^{\infty}\frac{E_1(p,t)}{E_0(p,t)}\dot{B}(p,t)dp$
where point denote derivative with respect to $\mu$ for $A(\mu,t)$ and $p$ for $B(p,t)$.
But I do not know where my mistakes are, maybe is there something liked with the chain rule that I do not consider ?.
Thanks for your help