How to Differentiate Bessel Functions

Title: How to Differentiate Bessel Functions Category: mathematics

Bessel functions have derivative formulas that connect neighboring orders. The exact signs depend on which family of Bessel functions you are differentiating. The formulas below assume a positive real argument $x>0$ and differentiation with respect to the argument $x$, not the order $\nu$.

Derivatives of the ordinary Bessel functions

For the Bessel functions of the first and second kinds, $J_\nu(x)$ and $Y_\nu(x)$, the derivative is the difference of the functions at adjacent orders:

$$\begin{aligned} J_\nu'(x) &= \frac{J_{\nu-1}(x)-J_{\nu+1}(x)}{2},\\ Y_\nu'(x) &= \frac{Y_{\nu-1}(x)-Y_{\nu+1}(x)}{2}. \end{aligned}$$

Equivalent forms follow from the recurrence relations:

$$\begin{aligned} J_\nu'(x) &= J_{\nu-1}(x)-\frac{\nu}{x}J_\nu(x) = \frac{\nu}{x}J_\nu(x)-J_{\nu+1}(x),\\ Y_\nu'(x) &= Y_{\nu-1}(x)-\frac{\nu}{x}Y_\nu(x) = \frac{\nu}{x}Y_\nu(x)-Y_{\nu+1}(x). \end{aligned}$$

The single-order forms involve $1/x$ and are stated for $x>0$. At $x=0$, use the relevant series to determine whether a finite derivative exists; in particular, $Y_\nu$ is generally singular there.

Derivatives of the modified Bessel functions

The modified Bessel functions have related formulas, but the signs differ. For the modified Bessel function of the first kind, $I_\nu$, the adjacent orders are added:

$$I_\nu'(x)=\frac{I_{\nu-1}(x)+I_{\nu+1}(x)}{2} =I_{\nu-1}(x)-\frac{\nu}{x}I_\nu(x) =I_{\nu+1}(x)+\frac{\nu}{x}I_\nu(x).$$

For the modified Bessel function of the second kind, $K_\nu$, the corresponding sum has a minus sign:

$$K_\nu'(x)=-\frac{K_{\nu-1}(x)+K_{\nu+1}(x)}{2} =-K_{\nu-1}(x)-\frac{\nu}{x}K_\nu(x) =-K_{\nu+1}(x)+\frac{\nu}{x}K_\nu(x).$$

A quick memory aid is that $J$ and $Y$ use a difference, while $I$ and $K$ use a sum; the derivative of $K$ has an overall minus sign in the adjacent-order formula.

Weighted derivative identities

If the function includes a power of $x$, differentiating the product directly may be unnecessary. For the ordinary Bessel functions,

$$\begin{aligned} \frac{d}{dx}\bigl(x^\nu J_\nu(x)\bigr)&=x^\nu J_{\nu-1}(x),\\ \frac{d}{dx}\bigl(x^{-\nu}J_\nu(x)\bigr)&=-x^{-\nu}J_{\nu+1}(x). \end{aligned}$$

The same identities hold with $J$ replaced by $Y$. For the modified functions, the corresponding identities are

$$\begin{aligned} \frac{d}{dx}\bigl(x^\nu I_\nu(x)\bigr)&=x^\nu I_{\nu-1}(x),\\ \frac{d}{dx}\bigl(x^{-\nu}I_\nu(x)\bigr)&=x^{-\nu}I_{\nu+1}(x),\\ \frac{d}{dx}\bigl(x^\nu K_\nu(x)\bigr)&=-x^\nu K_{\nu-1}(x),\\ \frac{d}{dx}\bigl(x^{-\nu}K_\nu(x)\bigr)&=-x^{-\nu}K_{\nu+1}(x). \end{aligned}$$

These are particularly useful when the Bessel function already appears multiplied by a power of its argument.

Worked example: differentiating a weighted Bessel function

Differentiate $f(x)=x^2J_2(x)$. Here the power of $x$ matches the order, so apply the first weighted identity with $\nu=2$:

$$f'(x)=\frac{d}{dx}\bigl(x^2J_2(x)\bigr)=x^2J_1(x).$$

You can check this using the product rule and the single-order formula. Since $J_2'(x)=J_1(x)-2J_2(x)/x$,

$$\begin{aligned} f'(x)&=2xJ_2(x)+x^2J_2'(x)\\ &=2xJ_2(x)+x^2\left(J_1(x)-\frac{2}{x}J_2(x)\right)\\ &=x^2J_1(x). \end{aligned}$$

The terms involving $J_2$ cancel, which is why the weighted identity is convenient.

Differentiating a Bessel function of a function

If the argument is itself a function, apply the chain rule. For example, if $h(x)=J_\nu(3x)$, then

$$h'(x)=3J_\nu'(3x) =\frac{3}{2}\left[J_{\nu-1}(3x)-J_{\nu+1}(3x)\right].$$

The factor of $3$ comes from differentiating the argument $3x$. More generally, for $J_\nu(g(x))$, multiply the derivative formula evaluated at $g(x)$ by $g'(x)$.

Common mistakes

• Confusing argument and order derivatives. A prime in these formulas means differentiation with respect to $x$. Differentiating with respect to $\nu$ is a different operation and is not given by these recurrence identities.
• Using an ordinary formula for a modified function. The adjacent-order signs for $I_\nu$ and $K_\nu$ differ from those for $J_\nu$ and $Y_\nu$.
• Forgetting the chain-rule factor. For $J_\nu(g(x))$, include $g'(x)$.
• Dropping a power-rule term. When differentiating $x^aJ_\nu(x)$ for arbitrary $a$, use the product rule unless a weighted identity applies. The simplest identity above requires the exponent to match the order (or its negative).