Can anybody explain the formula for Kinetic Energy?

To be specific, my real question is this: since the first derivative squared and the second derivative are generally not equal, why is it that we assert in undergrad phys that mechanical work done on a particle is equal to the kinetic energy gained?
Work = $\int \frac{d\; \vec{p}}{dt} \bullet d \vec{x} = m \int \frac{d^2 \vec{x}}{dt^2} \bullet d \vec{x}$.
But the kinetic energy is $\int \vec{p} \bullet d \vec{v} = m \int \frac{d \vec{x}}{dt} \bullet dfrac{d \vec{x}}{dt}$
It seems as though this is saying that the first derivative squared is equal to the second derivative of x with respect to t.
The same thing happens in the Schrodinger equation, when p is squared to set up the kinetic energy. The momentum operator in coordinate space is squared, and this is taken to be equivalent to the second derivative with respect to x, which is true when the function is the plane wave, but not in general. But, since all wave functions can be expanded in a fourier series or transform, we can get away with this seeming slip of the pen.
What is it about energy that, regardless of the functional dependence of the position with respect to time, we can unequivocally say that the first derivative squared is equal to the second derivative?

since the first derivative squared and the second derivative are generally not equal, why is it that we assert in undergrad phys that mechanical work done on a particle is equal to the kinetic energy gained?
Work = $\int \frac{d\; \vec{p}}{dt} \bullet d \vec{x} = m \int \frac{d^2 \vec{x}}{dt^2} \bullet d \vec{x}$.
But the kinetic energy is $\int \vec{p} \bullet d \vec{v} = m \int \frac{d \vec{x}}{dt} \bullet d \frac{d \vec{x}}{dt}$
It seems as though this is saying that the first derivative squared is equal to the second derivative of x with respect to t. The same thing happens in the Schrodinger equation
"Seems"? But you still haven't exactly proven such, unlike the "assertion of undergraduate physics" which is proven thus:

$$W =\int F\ dx =m \int \frac{d v}{dt}dx =m \int \frac{d x}{dt}\frac{d v}{dx}dx =m \int \frac{d \frac 1 2 v^2 }{dv}\frac{d v}{dx}dx =m \int d (\frac 1 2 v^2) =\big[ \frac 1 2 m v^2 \big] =\Delta KE$$

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