Parametric Derivative Derivation

1. Nov 20, 2005

How does one conclude that $$\frac{d^{2} y}{dx^{2}} = \frac{dy\'/dt}{dx/dt}$$?

Thanks

2. Nov 20, 2005

cronxeh

3. Nov 22, 2005

John_Doe

$$\frac{d^2 y}{dx^2} = \frac{d \dot{y'}}{d \dot{x}}$$
$$= \frac{dy'}{dx} = \frac{dy}{dx^2}$$

Which is clearly incorrect.

Last edited: Nov 22, 2005
4. Nov 22, 2005

HallsofIvy

Staff Emeritus
And, you will notice that what you wrote was incorrect. What is true is that $$\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ not the second derivative.
That follows from the chain rule.

5. Nov 22, 2005

I didn't write
. I wrote $$\frac{d^{2}y}{dx^{2}}= \frac{\frac{dy'}{dt}}{\frac{dx}{dt}}$$

6. Nov 24, 2005

John_Doe

Yes. Oops.
$$\frac{d^2 y}{dx^2} = \frac{d \dot{y'}}{d \dot{x}}$$
$$= \frac{dy'}{dx} = y''$$

7. Nov 25, 2005

HallsofIvy

Staff Emeritus
Then is y' the derivative with respect to x or t? If with respect to t, then your equation is wrong. If with respect to x, then you are back to the first derivative case.

8. Nov 25, 2005

John_Doe

$$y'' = \frac{d^2 y}{dx^2}$$