Derivative of x^n by Bessel Function of the First Kind of Order n of x

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Theorem

Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.


Then:

$\map {\dfrac \d {\d x} } {x^n \map {J_n} x} = x^n \map {J_{n - 1} } x$


Proof

\(\ds \map {\frac \d {\d x} } {x^n \map {J_n} x}\) \(=\) \(\ds n x^{n - 1} \map {J_n} x + x^n \map {J_n'} x\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds n x^{n - 1} \map {J_n} x - x^n \paren {\frac {\map {J_{n + 1} } x - \map {J_{n - 1} } x} 2}\) Recurrence Formula for Bessel Function of the First Kind
\(\ds \) \(=\) \(\ds n x^{n - 1} \map {J_n} x - x^n \paren {\frac {\frac {2 n} x \map {J_n} x - 2 \map {J_{n - 1} } x} 2}\) Recurrence Formula for Bessel Function of the First Kind
\(\ds \) \(=\) \(\ds n x^{n - 1} \map {J_n} x - x^n \paren {\frac n x \map {J_n} x - \map {J_{n - 1} } x}\)
\(\ds \) \(=\) \(\ds n x^{n - 1} \map {J_n} x - n x^{n - 1} \map {J_n} x + x^n \map {J_{n - 1} } x\)
\(\ds \) \(=\) \(\ds x^n \map {J_{n - 1} } x\)

$\blacksquare$


Sources