Derivative of x(t)?

  • #1
EastWindBreaks
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Homework Statement


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The Attempt at a Solution


I am trying to repair my rusty calculus. I don't see how du = dx*dt/dt, I know its chain rule, but I got (du/dx)*(dx/dt) instead of dxdt/dt, if I recall correctly, you cannot treat dt or dx as a variable, so they don't cancel out. so for (du/dx)(dx/dt) to become dxdt/dt, du/dx must equal to dt, which is not...
 

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Answers and Replies

  • #2
PeroK
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What the book has is a bit absurd. If ##u = x(t)##, then ##u## and ##x## are the same function. That's not really a substitution. Instead, you can use

##dx = \frac{dx}{dt} dt##

Directly.
 
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  • #3
EastWindBreaks
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What the book has is a bit absurd. If ##u = x(t)##, then ##u## and ##x## are the same function. That's not really a substitution. Instead, you can use

##dx = \frac{dx}{dt} dt##

Directly.
thank you, I see, so it doesn't matter if x is a function of t, we just integrate it regularly?
 
  • #4
PeroK
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thank you, I see, so it doesn't matter if x is a function of t, we just integrate it regularly?

If you have an integral in ##x##, you have an integral in ##x##. It doesn't matter that you can express ##x## as a function of another variable. In a way, you can always do that.
 
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