Notation of a Parametric Derivative

[Char. Limit]

Now first, as I'm sure you all know, for any two parametric equations x(t) and y(t), the slope of the curve thus generated is equal to the quotient (dy/dt)/(dx/dt). Also, we know that the concavity of the function (or change of slope, as I'm not sure if concavity is actually a correct term) is equal to the quotient (dy'/dt)/(dx/dt), where y'=dy/dx=(dy/dt)/(dx/dt). Now to my question... Is it allowed to represent dy'/dt as... I'll switch to latex now to make sure it comes out right...

$$\frac{d^2 y}{dx dt}$$

...this?

 

[Mark44]

Now first, as I'm sure you all know, for any two parametric equations x(t) and y(t), the slope of the curve thus generated is equal to the quotient (dy/dt)/(dx/dt). Also, we know that the concavity of the function (or change of slope, as I'm not sure if concavity is actually a correct term) is equal to the quotient (dy'/dt)/(dx/dt), where y'=dy/dx=(dy/dt)/(dx/dt). Now to my question... Is it allowed to represent dy'/dt as... I'll switch to latex now to make sure it comes out right...

$$\frac{d^2 y}{dx dt}$$

...this?

I'll take a stab at what I think you're trying to say.

You have a curve defined by the parametric equation x = x(t) and y = y(t). Assuming that these functions are differentiable, we can talk about
$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

Then
$$\frac{d^2y}{dx^2} = \frac{d}{dx}\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

$$= \frac{d}{dt} \left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right) \frac{dx}{dt}$$

 

[Unit]

$$\frac{d}{dx}\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

$$= \frac{d}{dt} \left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right) \frac{dx}{dt}$$

Actually, I think, by the chain rule, it should be

$$\frac{d}{dx} \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{d}{dt} \left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right) \frac{dt}{dx}$$

 

[Mark44]

You are correct, Unit.

 

[Char. Limit]

Actually, I'm just wondering if d/dt(dy/dx) can be represented by the notation in the OP in laTEX. It seems like an odd notation, with mixed total derivatives or something.

 

[Mark44]

AFAIK, no. For partials, yes, but I don't recall ever seeing ordinary derivatives stacked up like that.