# Calculation of spherical Bessel functions - meaning of $\left(\frac{1}{x}\frac{d}{dx}\right)^{l}$

I'm trying to understand the calculation of spherical Bessel functions in chapter four of Griffiths' Introduction to Quantum Mechanics (2nd ed, p142). He gives $$j_{2}\left(x\right)=\left(-x\right)^{2}\left(\frac{1}{x}\frac{d}{dx}\right)^{2}\frac{\sin x}{x}=x^{2}\left(\frac{1}{x}\frac{d}{dx}\right)\frac{x\cos x-\sin x}{x^{3}}$$

$$=\frac{3\sin x-3x\cos x-x^{2}\sin x}{x^{3}}.$$

I can't see how he arrives at this answer. I think my problem is the $\left(\frac{1}{x}\frac{d}{dx}\right)^{2}$ bit (the general term for $j_{l}\left(x\right)$ is $\left(\frac{1}{x}\frac{d}{dx}\right)^{l}$ ). I'm assuming this means $1/x^{2}$ multiplied by the second derivative of $\frac{\sin x}{x}$ but I make that $$\left(\frac{1}{x^{2}}\right)\left(-\frac{\sin x}{x}+\frac{2\sin x}{x^{3}}-\frac{2\cos x}{x^{2}}\right).$$

Any idea what I'm doing wrong?

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No. The term $\left(\frac1x \frac d{dx}\right)^2$ should be understood to mean the operator $$\left(\frac1x \frac d{dx}\right)^2=\frac1x \frac d{dx}\frac1x \frac d{dx}.$$ Otherwise one would simply say $\frac1{x^2} \frac {d^2}{dx^2}$. Check the derivation of the formula you use for the spherical Bessel functions to get a feel for why this must be the case.
Yes Peter, thats what you do. $\frac{1}{x}\frac{d}{dx} \frac{1}{x} \frac{d}{dx} \phi= \frac{1}{x}\frac{d}{dx} \frac{1}{x} \phi^\prime$. Mind that you need to apply of the product rule for derivations when you use the second $\frac{d}{dx}$ term –  elcojon Jan 15 '13 at 17:58