# The Derivative of Bessel Function of the Second Kind

Hello,

What is $$\frac{d}{dx}K_v\left(f(x)\right)=?$$

Thanks in advance

## Answers and Replies

Mute
Homework Helper
For $\nu$ not necessarily an integer, $C_{\nu}(y) = e^{\nu \pi i}K_{\nu}(y)$ satisfies the identity

$$2\frac{dC_\nu}{dy} = C_{\nu-1}(y) + C_{\nu + 1}(y)$$

Then let $y = f(x)$ and use the chain rule.

Identities like this can usually be found on wikipedia, and for something as studied and used as frequently as the Bessel Functions, are generally correct.

http://en.wikipedia.org/wiki/Bessel_function

For a more 'official' reference, see something like http://www.math.sfu.ca/~cbm/aands/page_437.htm (scans of a reference book).

For $\nu$ not necessarily an integer, $C_{\nu}(y) = e^{\nu \pi i}K_{\nu}(y)$ satisfies the identity

$$2\frac{dC_\nu}{dy} = C_{\nu-1}(y) + C_{\nu + 1}(y)$$

Then let $y = f(x)$ and use the chain rule.

Identities like this can usually be found on wikipedia, and for something as studied and used as frequently as the Bessel Functions, are generally correct.

http://en.wikipedia.org/wiki/Bessel_function

For a more 'official' reference, see something like http://www.math.sfu.ca/~cbm/aands/page_437.htm (scans of a reference book).

I know that
$$z\frac{d}{dz}K_v(z)+vK_v(z)=-zK_{v-1}(z)$$
but I was confused when we have more complicated arguments such as
$$z=\sqrt{x^2+x}$$.
But after your posting, I have now a simple method to move from simple to more complicated arguments. So, I can say the following:

$$\frac{f(z)}{f'(z)}\frac{d}{dz}K_v(f(z))+vK_v(f(z))=-f(z)K_{v-1}(f(z))$$

Am I right?

Best regards

Hello,

What is $$\frac{d}{dx}K_v\left(f(x)\right)=?$$

Thanks in advance

Do you get the answer?

Do you get the answer?

let $$y=f(x)$$ and then use the chain rule.

I know that
$$z\frac{d}{dz}K_v(z)+vK_v(z)=-zK_{v-1}(z)$$
but I was confused when we have more complicated arguments such as
$$z=\sqrt{x^2+x}$$.
But after your posting, I have now a simple method to move from simple to more complicated arguments. So, I can say the following:

$$\frac{f(z)}{f'(z)}\frac{d}{dz}K_v(f(z))+vK_v(f(z))=-f(z)K_{v-1}(f(z))$$

Am I right?

Best regards

i think this is right

let $$y=f(x)$$ and then use the chain rule.

thanks a lot