Chain rule problem

  • #1
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Summary:

Should be a simple chain rule problem I think.

Main Question or Discussion Point

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!
 

Answers and Replies

  • #2
PeroK
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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.
 

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