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The derivative of Bessel function of first kind (zero order, $J'_0$) is $-J_1$. What is the derivative of Bessel function of second kind (zero order, $Y'_0$)?

I could find $I'_0$ and $K'_0$, but not $Y'_0$.

Thanks in advance!

Felix B.
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js2003
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1 Answers1

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These derivatives are really easy to memorize

$$\frac d {dx}I_0(x)=+I_1(x)$$ $$\frac d {dx}J_0(x)=-J_1(x)$$ $$\frac d {dx}Y_0(x)=-Y_1(x)$$ $$\frac d {dx}K_0(x)=-K_1(x)$$

Now, for higher orders $$\frac d {dx}I_n(x)=+\frac{1}{2} (I_{n-1}(x)+I_{n+1}(x))$$ $$\frac d {dx}J_n(x)=+\frac{1}{2} (J_{n-1}(x)-J_{n+1}(x))$$ $$\frac d {dx}Y_n(x)=+\frac{1}{2} (Y_{n-1}(x)-Y_{n+1}(x))$$ $$\frac d {dx}K_n(x)=-\frac{1}{2} (K_{n-1}(x)+K_{n+1}(x))$$

Claude Leibovici
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  • does $n$ need to be a natural number or can it be an arbitrary positive number? How do you deal with $n\in(0,1)$? – Felix B. Feb 28 '23 at 09:35
  • https://functions.wolfram.com/Bessel-TypeFunctions/BesselK/20/01/02/ looks like this is true in general. And negative $n$ are valid. I just thought that it had to be positive for some reason – Felix B. Feb 28 '23 at 09:58