Calculus: parametric differentiation

[Uku]

Homework Statement

I need to know why:

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

$$y=y(t) , x=x(t)$$

The Attempt at a Solution

I know that:

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

From here:

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

And taking the second derivative with respect to x I get:

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

But how do the two equal?

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

 

[tiny-tim]

Hi Uku!

d/dt (dy/dx) is equal to d/dx (dy/dt) …

more generally, d²/dtdx = d²/dxdt

but simpler would be to use the chain rule:

d/dx (dy/dx) = d/dt (dy/dx) dt/dx

 

[Uku]

Hi! I find it odd how I am unable to grasp this.
Am I wrong when I write?

$$y_{xx}=\frac{d}{dx}(y_{x})=\frac{\frac{d}{dt}(y_{x})}{\frac{dx}{dt}}=<br /> =\frac{d}{dt}(\frac{dx}{dt})^{-1}(y_{x})=\frac{d}{dx}(y_{x})$$

I'm sort of trying the backward motion here to get the point of my initial problem.

Thanks,
Uku

 

[tiny-tim]

Hi Uku!

Sorry, I don't understand your d/dt (dx/dt)^-1 (y_x)

 

[FLUndergrad]

Hi Uku!

d/dt (dy/dx) is equal to d/dx (dy/dt) …

more generally, d²/dtdx = d²/dxdt

but simpler would be to use the chain rule:

d/dx (dy/dx) = d/dt (dy/dx) dt/dx

Oh wow I never thought if it that way, that is awesome. I think that just cleared up a LOT of ambiguity I had about the whole d?/d? notation system thing, which to be honest I never really understood that well to begin with.

 

[Uku]

Thanks for the replies!