I would like help showing the following is true:
$$\frac{d}{dx}[x^{-\alpha}J_{\alpha}(x)] = -x^{-\alpha}J_{\alpha+1}(x).$$
I can show $\frac{d}{dx}[x^{\alpha}J_{\alpha}(x)] = x^{\alpha}J_{\alpha-1}(x)$, however I am having a hard time extending this to the above.
Lets see the following recurrence relation between the Bessel function:
$$J_{n-1}(x)+J_{n+1}(x)=\displaystyle\frac{2n}{x} J_n(x)$$
$$J_{n-1}(x)-J_{n+1}(x)=2J'_n(x)$$
Adding these equations produces
$$J_{n-1}(x)=\displaystyle\frac{n}{x}J_n(x)+J'_n(x) $$
if you multiply by $x^n$ and play a little with the result you can get
$$\displaystyle\frac{d}{dx}[x^nJ_n(x)]=x^nJ_{n-1}(x)$$
in a similar way, if you subtract the equation, produces
$$J_{n+1}(x)=\displaystyle\frac{n}{x}J_n(x)-J'_n(x)$$
multiplying by $x^{-n}$ and playing with the equations, we obtain
$$\displaystyle\frac{d}{dx}[x^{-n}J_n(x)]=-x^{-n}J_{n+1}(x)$$
You can supply the details.