Parametric Derivative Derivation

[courtrigrad]

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

Thanks

 

[cronxeh]

Well if you googled "parametric derivative" you would've stumbled on your answer in about 3 seconds:
http://www.mathwords.com/p/parametric_derivative_formulas.htm

 

[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.

 

[HallsofIvy]

Well if you googled "parametric derivative" you would've stumbled on your answer in about 3 seconds:
http://www.mathwords.com/p/parametric_derivative_formulas.htm

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.

 

[courtrigrad]

I didn't write

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

. I wrote $$\frac{d^{2}y}{dx^{2}}= \frac{\frac{dy'}{dt}}{\frac{dx}{dt}}$$

 

[John_Doe]

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

 

[HallsofIvy]

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.

 

[John_Doe]

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