Deriving the Work-Energy Theorem: A Calculus Refresher

[nothing123]

It's been a while since I've taken calculus. I was going through the derivation of the work-energy theorem and came across this: dv/dt = (dv/dx)(dx/dt) which is supposed to be a result of the chain rule. Anyone care to explain and please simplify it as much as possible.

Thanks!

 

[nicksauce]

The chain rule is

f(g(x))' = f'(g(x))g'(x)

Now replace f with v, g with x, x with t,the first ' with d/dt, the second ' with d/dx (since f is a function of g aka x), and the third ' with d/dt and voila you have
dv/dt = (dv/dx)(dx/dt)

 

[tiny-tim]

Hi nothing123!

If v is a function of x only, and x is a function of t only, then if you increase t by a small amount ∆t, then x increases by a small amount ∆x = (dx/dt)∆t. (1)

But v also increases, by a small amount ∆v = (dv/dx)∆x. (2)

So, combining (1) and (2):

∆v = (dv/dx)∆x = (dv/dx)(dx/dt)∆t. ​

 

[nothing123]

Great, thanks for your help guys.