# Bessel functions of the first kind

1. Jul 5, 2010

### John 123

1. The problem statement, all variables and given/known data
Can anyone tell me if:
$$\frac{d}{dx}J_k(ax)=aJ'_k(x)$$
where a is a real positive constant and
$$J_k(x)$$
is the Bessel function of the first kind.
Regards
John

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 5, 2010

### Char. Limit

I believe that the answer to that is no. I believe that...

$$\frac{d}{dx}J_k(ax) = a J_k^{'}(ax)$$

3. Jul 5, 2010

### John 123

But doesn't
$$J_k(ax)=aJ_k(x)?$$

4. Jul 5, 2010

### Char. Limit

Sorry, but that is not the case.

5. Jul 5, 2010

### John 123

Then that is my error.
To find the derivative of
$$J_k(ax)$$
presumably one has to differentiate the series expansion?
John

6. Jul 5, 2010

### Dick

You can express the derivative of a bessel function in terms of other bessel functions. See the "Selected identities" list at the end of http://en.wikipedia.org/wiki/Bessel_function

7. Jul 5, 2010

### John 123

Hi Dick
The question I am asking derives from proving the integral property of Bessel functions of the first kind. This amounts to showing the orthogonal properties of Bessel functions.
Part way through the proof I need to show that:
$$u'_1(x)=r_ix^(\frac{1}{2})J'_k(r_ix)+\frac{1}{2}x^{\frac{-1}{2}}J_(r_ix)$$
where
$$r_i$$
is a distinct positive zero of
$$J_k(x)$$

8. Jul 5, 2010

### John 123

Hi Dick
I need to show that if:
$$u(x)=x^{\frac{1}{2}}J_k(r_ix)$$
Then:
$$u'=r_ix^{\frac{1}{2}}J'_k(r_ix)+\frac{1}{2}x^{\frac{-1}{2}}J_k(r_ix)$$
This, of course, uses product rule but the derivative of:
$$J_k(r_ix)$$
Incidentally
$$r_i$$
is the ith distinct positive zero of
$$J_k(x)$$
This is part of proving the orthogonality of Bessel functions
$$J_k(r_ix)$$
with respect to the weight function x.
John

9. Jul 5, 2010

### Dick

It's just the chain rule. If h(x)=f(a*x) then h'(x)=f'(a*x)*a. It's a special case of the general chain rule, if h(x)=f(g(x)) then h'(x)=f'(g(x))*g'(x). It's just like saying the derivative of sin(2*x) is cos(2*x)*2.

10. Jul 5, 2010

Thanks Dick
John