we already know that the Bessel series expansions is
$$ 
J\_n(\lambda r)=({\lambda r \over 2})^n \sum\_m^{\infty}({\lambda r \over 2})^{2m} \frac{(-1)^m}{m!(n+m)!}$$
we want to prove that
$$\frac{d}{dr}J\_0(\lambda r)= -\lambda J\_1(\lambda r)$$
we first notice that
$$ \frac{d}{dr}J\_0(\lambda r)= \lambda J\_0^{'}(\lambda r)$$
so the Eq.(2)becomes
$$\lambda J\_0^{'}(\lambda r)=-\lambda J\_1(\lambda r)$$
In other words we just need to prove that
$$J\_0^{'}(\lambda r)+J\_1(\lambda r)=0$$
obviously
$$J\_0(\lambda r)=\sum\_m^{\infty}({\lambda r \over 2})^{2m} \frac{(-1)^m}{m!m!} $$
so
$$J\_0^{'}(\lambda r)=2m\sum\_m^{\infty}({\lambda r \over 2})^{2m-1} \frac{(-1)^m}{m!m!}$$
moreover
$$J\_1(\lambda r)=({\lambda r \over 2})^1\sum\_m^{\infty}({\lambda r \over 2})^{2m} \frac{(-1)^m}{m!(1+m)!}$$
so that
$$J\_0^{'}(\lambda r)+J\_1(\lambda r)=2m\sum\_m^{\infty}({\lambda r \over 2})^{2m-1} \frac{(-1)^m}{m!m!}+({\lambda r \over 2})^1\sum\_m^{\infty}({\lambda r \over 2})^{2m} \frac{(-1)^m}{m!(1+m)!}$$
after some simplification I notice I just need to prove that
$$ (\frac{\lambda r}{2})^2 \frac{1}{m+1}+2m =0 $$
and I am stuck at here.
I am just wondering, where am I wrong?