Deriving the work-energy theorem step

[Jonnyb42]

This is probably more of a math question, but it pertains to physics.

Today in Classical Mechanics the professor was showing the following,

F . dr = (MV₂^2)/2 - (MV₁^2)/2

but in great length, one of the steps along the way I did not understand how he did it... it was this:

screw latex, if someone could please post a link to somewhere on the forums that shows how to use latex properly, I can't use it to save my life...
here is the step with good old microsoft:

[PLAIN]http://mynqa.com/Cargo/unk.bmp

My questions are, where did the number 2 come from in the denominator? Also, how did he bring in the velocity vector into the derivative like that?

Thanks for any help.

 

[gabbagabbahey]

screw latex, if someone could please post a link to somewhere on the forums that shows how to use latex properly, I can't use it to save my life...

See my sig

here is the step with good old microsoft:

[PLAIN]http://mynqa.com/Cargo/unk.bmp

My questions are, where did the number 2 come from in the denominator? Also, how did he bring in the velocity vector into the derivative like that?

Thanks for any help.

Just apply the product rule to the right-hand side of the equation...

 

[mathman]

It is essentially the vector equivalent of the derivative of a square.

d/dt(x²)=2xdx/dt