# Homework Help: Partial derivatives question

1. May 22, 2006

### Luminous Blob

I'm trying to follow a derivation in given in a text book. Part of this derivation goes like this:

$$\frac{d}{ds}\left(\frac{1}{c}\frac{dx}{ds}\right)=c\left(\frac{\partial^2\tau}{\partial x^2}\frac{\partial \tau}{\partial x} + \frac{\partial^2\tau}{\partial x \partial y}\frac{\partial \tau}{\partial y}\right)$$
$$=\frac{c}{2}\frac{\partial}{\partial x}\left[\left(\frac{\partial \tau}{\partial x}\right )^2 + \left (\frac {\partial \tau}{\partial y}\right )^2 \right]$$

I worked through that and came up with the same answer, but without the factor of 1/2. Can anyone tell me where it comes from?

Last edited: May 22, 2006
2. May 22, 2006

### J77

Differentiate out those last terms...

d/dx(d tau/dx)^2=2d tau/dx * (d^2 tau/dx^2)

3. May 22, 2006

### Luminous Blob

Ah, gotcha! Thanks :)

4. May 22, 2006

### Luminous Blob

Hang on, after looking at it a bit more I'm not so sure...wouldn't that give you a factor of 2 out the front rather than 1/2?

5. May 23, 2006

### George Jones

Staff Emeritus
Doesn't performing the differentiation in the bottom line result in the top line, since the 2 cancels the 1/2?

If it does, then isn't everything OK?

Regards,
George

6. May 24, 2006

### J77

Yep - the '2' in my previous post cancels with the '2' of Blob's last term, giving the middle term...

7. May 24, 2006

### Luminous Blob

Haha, I see now...as you may have noticed, I'm not exactly the sharpest tool in the shed :)

Thanks again.