Solving Calculus: Derivative of x(t)

[EastWindBreaks]

Homework Statement

Homework Equations

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...

 

[PeroK]

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.

 

[EastWindBreaks]

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?

 

[PeroK]

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.