Bessel function $J_{3\over 2}(x)$ How can i find $J_{3\over 2}(x)$ and $J_{5\over 2}(x)$ by use this formula :
$$J_{p+{\frac{1}{2}}}(x)=\left(\frac{2}{\pi}\right)^{\frac{1}{2}} \cdot(-1)^p x^{p+{\frac{1}{2}}}  
\left(\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\right)^p\left(\frac{\sin x}{x}\right)$$
I know if i want to find $J_{3\over 2}(x)$ substitute by $p=1$ and if I want to find $J_{5\over 2}(x)$ substitute by $p=2$ 
But i don't know how can i find $\left(\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\right)^2$
 A: For $p = 2$, the term $\displaystyle \left(\frac{1}{x} \frac{d}{dx}\right)^2$ should be understood as the operator
$$\displaystyle \left(\frac{1}{x} \frac{d}{dx}\right)^2=\frac{1}{x} \frac{d}{dx}\frac1x \frac d{dx}.$$
So for $p=2$
$$
\begin{align}
J_{{\frac{5}{2}}}(x)&=\left(\frac{2}{\pi}\right)^{\frac{1}{2}} \cdot(-1)^2 x^{{\frac{5}{2}}}  
\left(\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\right)^2\left(\frac{\sin x}{x}\right)\\
&=\sqrt\frac{2}{\pi}  x^{{\frac{5}{2}}}  
\left[\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\left(\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\frac{\sin x}{x}\right)\right]\\
&=\sqrt\frac{2}{\pi}  x^{{\frac{5}{2}}}  
\left[\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\left(\frac{1}{x}\frac{x \cos x-\sin x}{x^2}\right)\right]\\
&=\sqrt\frac{2}{\pi}  x^{{\frac{5}{2}}}  
\left[\frac{1}{x}\left(-\frac{3 x \cos x+(x^2-3) \sin x}{x^4}\right)\right]\\
&=-\sqrt\frac{2}{\pi}  x^{\frac{5}{2}}  
\frac{3 x \cos x+(x^2-3) \sin x}{x^5}\\
&=-\sqrt\frac{2}{\pi}    
\frac{3 x \cos x+(x^2-3) \sin x}{x^{5/2}}
\end{align}
$$
Check the result with WolframAlpha (alternate forms).
