Bessel Function's Recursive Relations
Theorem
$$ J_{\nu}(x)=\sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n+1) \Gamma (n+\nu+1)} \left(\frac{x}{2} \right)^{2n+\nu} \tag{1} $$
The function above is called the first kind Bessel function of order $\nu$. The first kind Bessel function $J_{\nu}(x)$ satisfies the following equation.
$$ \begin{align*} \frac{d}{dx}[x^{\nu} J_{\nu}(x)] &= x^{\nu}J_{\nu-1}(x) \tag{a} \\ \frac{d}{dx}[x^{-\nu}J_{\nu}(x)] &= -x^{-\nu}J_{\nu+1}(x) \tag{b} \\ J_{\nu-1}(x)+J_{\nu+1}(x) &= \frac{2\nu}{x}J_{\nu}(x) \tag{c} \\ J_{\nu-1}(x)-J_{\nu+1}(x) &= 2J^{\prime}_{\nu}(x) \tag{d} \\ J_{\nu}^{\prime}(x) = -\frac{\nu}{x}J_{\nu}(x)+J_{\nu-1}(x) &= \frac{\nu}{x}J_{\nu}(x)-J_{\nu+1}(x) \tag{e} \end{align*} $$
Proof
$(a)$
By multiplying $x^{\nu}$ to $(1)$ and then differentiating, it can be easily obtained.
$$ \begin{align*} \frac{d}{dx}[x^{\nu} J_{\nu}(x)] &= \frac{d}{dx} \left[ x^{\nu} \sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n+1) \Gamma (n+\nu+1)} \left(\frac{x}{2} \right)^{2n+\nu} \right) \\ &= \frac{d}{dx} \sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n+1) \Gamma (n+\nu+1)} \frac{x^{2n+2\nu}}{2^{2n+\nu}} \\ &= \sum \limits_{n=0}^{\infty} \frac{(-1)^{n}(2n+2\nu)}{\Gamma (n+1) \Gamma (n+\nu+1)} \frac{x^{2n+2\nu-1}}{2^{2n+\nu}} \tag{2} \end{align*} $$
Since the gamma function satisfies the relation $\Gamma (n+\nu+1)=(n+\nu)\Gamma (n+\nu)$, simplifying the $2(n+\nu)$ in the denominator of $(2)$ yields
$$ \begin{align*} \frac{d}{dx}[x^{\nu} J_{\nu}(x)] &= \sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n+1) \Gamma (n+\nu)} \frac{x^{2n+2\nu-1}}{2^{2n+\nu-1}} \\ &= x^{\nu}\sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n+1) \Gamma (n+\nu)} \frac{x^{2n+\nu-1}}{2^{2n+\nu-1}} \\ &= x^{\nu}J_{\nu-1}(x) \end{align*} $$
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$(b)$
The general approach to the proof is the same as $(a)$, but there are subtle parts that are easy to miss, so it is explained without omission. By multiplying $x^{-\nu}$ to $(1)$ like $(a)$ and then differentiating
$$ \begin{align*} \frac{d}{dx}[x^{-\nu} J_{\nu}(x)] &= \frac{d}{dx}\sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n+1) \Gamma (n+\nu+1)} \frac{x^{2n}}{2^{2n+\nu}} \\ &= \sum \limits_{n=0}^{\infty} \frac{(-1)^{n}2n }{\Gamma (n+1) \Gamma (n+\nu+1)} \frac{x^{2n-1}}{2^{2n+\nu}} \end{align*} $$
Since it is $\Gamma (n+1)=n\Gamma (n)$, after simplifying the $2n$ in the denominator and adjusting the order of $\frac{x}{2}$
$$ \begin{align*} \frac{d}{dx}[x^{-\nu} J_{\nu}(x)] &= x^{-\nu}\sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n) \Gamma (n+\nu+1)} \frac{x^{2n+\nu-1}}{2^{2n+\nu-1}} \end{align*} $$
Here, let’s substitute the index with $n=k+1$. Then
$$ \begin{align*} \frac{d}{dx}[x^{-\nu} J_{\nu}(x)] &= x^{-\nu}\sum \limits_{k=-1}^{\infty} \frac{(-1)^{k+1} }{\Gamma (k+1) \Gamma (k+\nu+2)} \frac{x^{2k+\nu+1}}{2^{2k+\nu+1}} \end{align*} $$
If it is $k=-1$, the denominator diverges to $\Gamma (k+1)=\Gamma (0)=\infty$, so it is $0$. Therefore, it is okay to start the index from $k=0$.
$$ \begin{align*} \frac{d}{dx}[x^{-\nu} J_{\nu}(x)] &= x^{-\nu}\sum \limits_{k=0}^{\infty} \frac{(-1)^{k+1} }{\Gamma (k+1) \Gamma (k+\nu+2)} \frac{x^{2k+\nu+1}}{2^{2k+\nu+1}} \\ &= -x^{-\nu}J_{\nu+1}(x) \end{align*} $$
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$(c)$, $(d)$
By organizing $(a)$ for $J_{\nu-1}(x)$ and organizing $(b)$ for $J_{\nu+1}(x)$, then adding and subtracting, it can be immediately obtained.
$$ \begin{align*} J_{\nu-1}(x) &= x^{-\nu} \frac{d}{dx}\left[x^{\nu} J_{\nu}(x)\right] \\ &= x^{-\nu} \nu x^{\nu-1}J_{\nu}(x)+x^{-\nu}x^{\nu}J_{\nu}^{\prime}(x) \\ &= \nu x^{-1}J_{\nu}(x) + J_{\nu}^{\prime}(x) \end{align*} \tag{3} $$
$$ \begin{align*} J_{\nu+1}(x) &= -x^{\nu} \frac{d}{dx}\left[x^{-\nu} J_{\nu}(x) \right] \\ &= -x^{\nu}(-\nu) x^{-\nu-1}J_{\nu}(x)-x^{\nu}x^{-\nu}J_{\nu}^{\prime}(x) \\ &= \nu x^{-1}J_{\nu}(x) - J_{\nu}^{\prime}(x) \end{align*} \tag{4} $$
If you calculate $(3)+(4)$
$$ J_{\nu-1}+J_{\nu+1}(x)=\frac{2\nu}{x}J_{\nu}(x) $$
If you calculate $(3)-(4)$
$$ J_{\nu-1}-J_{\nu+1}(x)=2J_{\nu}^{\prime}(x) $$
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$(e)$
By expanding and organizing the left sides of $(a)$ and $(b)$, it can be easily obtained. First, by expanding the left side of $(a)$ and organizing for $J_{\nu}^{\prime}(x)$
$$ \begin{align*} \frac{d}{dx}[x^{\nu} J_{\nu}(x)] &=x^{\nu}J_{\nu-1}(x) \\ \nu x^{\nu-1}J_{\nu}(x) + x^{\nu} J_{\nu}^{\prime}(x) &= x^{\nu}J_{\nu-1}(x) \\ J_{\nu}^{\prime}(x) &= -\frac{\nu}{x}J_{\nu}(x)+J_{\nu-1}(x) \end{align*} $$
The same process is done for $(b)$
$$ \begin{align*} \frac{d}{dx}[x^{-\nu} J_{\nu}(x)] &=-x^{-\nu}J_{\nu+1}(x) \\ -\nu x^{-\nu-1}J_{\nu}(x) + x^{-\nu} J_{\nu}^{\prime}(x) &=-x^{-\nu}J_{\nu+1}(x) \\ J_{\nu}^{\prime}(x) &= \frac{\nu}{x}J_{\nu}(x)-J_{\nu+1}(x) \end{align*} $$
Therefore
$$ J_{\nu}^{\prime}(x)=-\frac{\nu}{x}J_{\nu}(x)+J_{\nu-1}(x)=\frac{\nu}{x}J_{\nu}(x)-J_{\nu+1}(x) $$
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