I How is the derivative of y(x,t) derived in quantum mechanics?

[strawman]

TL;DR Summary

Should be a simple chain rule problem I think.

y(x,t) = 1/2 h(x-vt) + 1/2 h(x+vt)
This is from the textbook "quantum mechancs" by Rae.
The derivative is given as dy/dt = -v1/2 h(x-vt) + v1/2 h(x+vt)
I'm not quite sure how this is? If I use the chain rule and set the function h(x-vt) = u
Then by dy/dt = dy/du x du/dt I will get (for the first part):

dy/du = 1/2
du/dt = -v
dy/dt = -1/2 v

I've obviously misunderstood this somewhere, being a bit rusty on my calculus. Where have I gone wrong?
Thanks!

 

[PeroK]

Summary: Should be a simple chain rule problem I think.

y(x,t) = 1/2 h(x-vt) + 1/2 h(x+vt)
This is from the textbook "quantum mechancs" by Rae.
The derivative is given as dy/dt = -v1/2 h(x-vt) + v1/2 h(x+vt)
I'm not quite sure how this is? If I use the chain rule and set the function h(x-vt) = u
Then by dy/dt = dy/du x du/dt I will get (for the first part):

dy/du = 1/2
du/dt = -v
dy/dt = -1/2 v

I've obviously misunderstood this somewhere, being a bit rusty on my calculus. Where have I gone wrong?
Thanks!

First, in the correct answer, those should be derivatives of $h$.

Second, your substitution $u = h$ changes nothing. You just replace $h$ by $u$. The two functions are identical.

What you can do is let $u = x - vt$. Then:

$y(x, t) = h(u(x, t)) + \dots$

It's that function composition to which the chain rule applies:

$\frac{dy}{dt} = h'(u)\frac{du}{dt} + \dots$

Note that it's $h'$ there.