Time Derivative: How Does 2x Differ from x² Differ?

[delve]

I'm wondering, how does 2 multiplied by the first and second time derivatives of x equal the time derivative of the time derivative of x squared. Thanks.

 

[Born2bwire]

Ummm... You mean,
$$2\frac{d^2}{dt^2} f(t) = \frac{d}{dt} \left(\frac{d}{dt} f(t) \right)^2$$
This would be done by the chain rule. That is,
$$\frac{d}{dx} (f(x))^2 = 2f(x)\frac{d}{dx}f(x)$$
In this sense, we take the two from the power of d/dt f(t) and take that as a coefficient, reduce the power by one, and then take the time derivative of d/dt f(t). I originally interpreted your question as
$$2\frac{d}{dt}\frac{d^2}{dt^2} f(t) = \frac{d}{dt} \left(\frac{d}{dt} f(t) \right)^2$$
but I do not feel this expression is true.

EDIT: Ok, let's fix this.

Ummm... You mean,
$$2\frac{d}{dt}f(t)\frac{d^2}{dt^2} f(t) = \frac{d}{dt} \left(\frac{d}{dt} f(t) \right)^2$$
This would be done by the chain rule. That is,
$$\frac{d}{dx} (f(x))^2 = 2f(x)\frac{d}{dx}f(x)$$
In this sense, we take the two from the power of d/dt f(t) and take that as a coefficient, reduce the power by one, and then take the time derivative of d/dt f(t).

 

[nicksauce]

Let v = dx/dt and a = dv/dt. Then I believe he means, why does 2va = d/dt (v^2)?

It immediately follows from the chain rule:

d/dt(v^2) = 2v*d/dt(v) = 2va

 

[Born2bwire]

Let v = dx/dt and a = dv/dt. Then I believe he means, why does 2va = d/dt (v^2)?

It immediately follows from the chain rule:

d/dt(v^2) = 2v*d/dt(v) = 2va

Ahh gotcha, I missed an f in there when I wrote it out on paper.

$$2\frac{d}{dt}f(t)\frac{d^2}{dt^2} f(t) = \frac{d}{dt} \left(\frac{d}{dt} f(t) \right)^2$$

Of course that would make my original statement wrong too...

I need more coffee.

 

[delve]

Awesome, thanks a lot for the help. I appreciate it.