I'm studying generalised Fourier series and part of proving an inner product I'm meant to show that $$\frac12(\mathrm J'_m(j_{mk}))^2= \frac12(\mathrm J_{m+1}(j_{mk}))^2 $$ Where $\mathrm J_{m}(x)$ is the Bessel function of the first kind and {$j_{mk}$} are the roots of the Bessel function of the first kind, $k=1,2,3,..$.
I know of two identities:
(i) $mJ_{m}(x)=\frac12x(J_{m-1}(x)+J_{m+1}(x))$
(ii) $J'_m(x)=\frac12(J_{m-1}(x)-J_{m+1}(x))$
How can I get from these two to the desired result? I've tried playing around with these identities but I can't seem to get it.
If $j_{mk}$ is a zero of $J_m$ then (i) gives: $$ J_{m-1}(j_{mk}) = -J_{m+1}(j_{mk})\tag{1}$$ and if we plug in this identity into (ii) we get: $$ J_m'(j_{mk}) = -J_{m+1}(j_{mk})\tag{2} $$ so the claim follows.