Anyone know how to take the 2nd derivative of a spherical bessel function? i'm trying to take the 2nd derivative of a spherical bessel function. So far i've found these recursion relations that are useful from:
http://dlmf.nist.gov/10.51
$(1) j'_{n}(x) = -j_{n+1}(x) + \frac{n}{x}j_{n}(x)$
$(2) j'_{n}(x) = j_{n-1}(x) - \frac{n+1}{x}j_{n}(x)$
What i've attempted is something like this:
So starting with (1): $j'_{n}(x) = -j_{n+1}(x) + \frac{n}{x}j_{n}(x)$
I'm wondering if you can say : $j''_{n}(x) = -j'_{n+1}(x) +D[\frac{n}{x}j_{n}(x)]$ where $D[\frac{n}{x}j_{n}(x)]$ is some product rule differentiation that i'm not sure how you would apply.
If this is valid I'm also wondering if you can use (2) on the $-j'_{n+1}(x)$ term so that:
$-j'_{n+1}(x) = j_{(n+1)-1}(x) - \frac{(n+1)+1}{x}j_{n+1}(x)= j_{n}(x)-\frac{n+2}{x}j_{n+1}(x)$
and would the $D[\frac{n}{x}j_{n}(x)]$ term just be a simple product so that :
$D[\frac{n}{x}j_{n}(x)] = \frac{n}{x}j'_{n}(x) + \frac{n}{-x^2}j_{n}(x)$
 A: For the sake of completeness, I will restate the two recurrence relations given in the original post:
$$j_n(x)=-j_{n+1}(x)+\frac{n}{x}j_n(x), \hspace{2cm} \text{(1)}$$
$$j_n(x)=j_{n-1}(x)-\frac{n+1}{x}j_n(x), \hspace{2cm} \text{(2)}$$ 
We can use (2) to derive the following relation (setting $n=n+1$),
$$j_{n+1}(x)=j_{n}(x)-\frac{n+2}{x}j_{n+1}(x).\hspace{2cm} \text{(2*)}$$
this will be useful in the following calculation. 
We can apply differentiation to both sides of (1), I will omit writing the argument of the spherical Bessel functions for brevity,
$$\frac{d}{dx}j'_n=-\frac{d}{dx}j_{n+1}+n\frac{d}{dx}\Big(\frac{j_n}{x}\Big).$$
Applying the derivatives and using the quotient rule on the second term on the RHS gives
$$j''_n=-j'_{n+1}+n\Big(\frac{xj'_n-j_n}{x^2}\Big).$$
If we multiply by $x^2$ and move all terms to one side, we have
$$x^2j''_n+x^2j'_{n+1}-nxj'_n+nj_n=0.$$
Now, if we apply (2*) to $j'_{n+1}$ and (1) to $j'_n$, we arrive at the following
$$x^2j''_n+x^2\Big[j_n-\frac{n+2}{x}j_{n+1}\Big]-nx\Big[-j_{n+1}+\frac{n}{x}j_n\Big]+nj_n=0.$$
This can be rearranged to give our final result, the second derivative of the spherical Bessel function as a function of spherical Bessel functions:
$$j''_n(x)=\frac{1}{x^2}\Big[(n^2-n-x^2)j_n(x)+2xj_{n+1}(x)\Big].$$
Note than there's other ways to write this by using the opposite recurrence relations etc.
A: Yes, you can differentiate the recursion equations.  Alternatively, you know from the original equation that
$$x^2 j_n^{\prime\prime} + 2 x j_n^\prime + \left[ x^2-n(n+1) \right] j_n = 0$$ 
So, use this together with your recursion equation for $j_n^\prime$.  
