Differentiation of Bessel function
1,135
The answer is $$\frac{n^2x^2}{x} J_n(x)$$
You can get it using two properties of Bessel function:
$$J'_n(x) = \frac12(J_{n1}(x)J_{n+1}(x))$$
$$J_n(x) = \frac{x}{2n}(J_{n1}(x)+J_{n+1}(x))$$
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susovan0010
Updated on August 01, 2022Comments

susovan0010 over 1 year
How can I represent the following differentiation in terms of $J_n$? The equation is: $\frac{\partial}{\partial x}[xJ_n'(x)]$