Parametric Derivative Derivation (1 Viewer)

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

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

Thanks

Gold Member

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:

HallsofIvy

cronxeh said:
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.

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

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

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving