Differentiation of Bessel function

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The answer is $$\frac{n^2-x^2}{x} J_n(x)$$

You can get it using two properties of Bessel function:

$$J'_n(x) = \frac12(J_{n-1}(x)-J_{n+1}(x))$$

$$J_n(x) = \frac{x}{2n}(J_{n-1}(x)+J_{n+1}(x))$$

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susovan0010
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Updated on August 01, 2022

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  • susovan0010
    susovan0010 over 1 year

    How can I represent the following differentiation in terms of $J_n$? The equation is: $\frac{\partial}{\partial x}[xJ_n'(x)]$