# Square or second derivative?

1. Feb 9, 2014

### mrmojorizing

Hi, if oyu look at question 16b in the link below in order to get the second derivative wrt to t they take the square of the first derivative. I don't get it, how does multiplying the first derivative by itself get you the second derivative?

http://stuff.mit.edu/afs/athena/course/8/8.20/www/sols/sol1.pdf

Also, regarding the same question:

I am familiar with the chain rule: if y=f(g(t,x),h(t,x)) then dy/dt=dy/dg*dg/dt+dy/dh*dh/dt
To show that an equation is invariant under a galiliean transform, it’s partially necessary to show that the equation takes the same form both for x and for x’=x-v(T). So if you have a wave equation for E which applies for x, and t, you want to show that the wave equation, with all of its first and second derivatives also applies for x’ and t’.

For example if you look at question 16 b , they ask to show that the wave equation is not invariant under Galilean transforms. What I don’t understand is in this question why are they taking the derivative of E with respect to x and t, rather than with respect to x’ and t’. We already know the wave equation takes the correct form for x and t. We want to show that it doesn’t take the correct form for x’ and t’, so then why start off taking the derivative with respect to x and t, and muck about using the chain rule rather than taking the derivative with respect to x’, and t’ (which is what you’re really interested in).

Thanks

2. Feb 10, 2014

### Sunfire

What I am getting here is something physicists do when they write the derivatives in "operator form"

$(\frac{\partial}{\partial t'}- v \frac{\partial}{\partial x'})(\frac{\partial}{\partial t'}- v \frac{\partial}{\partial x'}) =$
$= (\frac{\partial}{\partial t'}- v \frac{\partial}{\partial x'})^2 =$
$=(\frac{{\partial}^2}{{\partial t'}^2}- 2v \frac{\partial}{\partial t'}\frac{\partial}{\partial x'} + v^2 \frac{{\partial}^2}{{\partial x'}^2})=$
$=(\frac{{\partial}^2}{{\partial t'}^2}- 2v \frac{{\partial}^2}{\partial t' \partial x'} + v^2 \frac{{\partial}^2}{{\partial} x'^2})$

Does this help a bit?

3. Feb 10, 2014