# Parametric equation confusion of d2y/dx2

1. Apr 1, 2012

### Eats Dirt

x=t2 and y=t3-3t, find dy/dx and d2y/dx2

I understand how to get to dy/dx but an confused on how to get to d2y/dx2 can someone please explain in depth, I know the formula is (d/dt dy/dx)/(dx/dt) I dont understand where d/dt comes from and what it is. why do we not just derive it twice? I really do not understand the 2nd part at all.

2. Apr 1, 2012

### Office_Shredder

Staff Emeritus
It's just several applications of the chain rule

$$\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = \left(\frac{dy}{dt} / \frac{dx}{dt} \right)$$
from what your OP says you're familiar with this calculation. We're just going to do it again
$$\frac{d}{dx} \left( \frac{dy}{dx} \right)$$$$= \frac{d}{dt} \left( \frac{dy}{dx} \right) \frac{dt}{dx}$$$$= \frac{d}{dt} \left( \frac{dy}{dt}/\frac{dx}{dt} right) / \frac{dx}{dt}$$

Last edited by a moderator: Apr 2, 2012
3. Apr 1, 2012

### Eats Dirt

Im sorry but i still have no idea : \.

4. Apr 1, 2012

### jppike

It will perhaps become more clear if you switch your notation slightly. How about we denote dy/dx instead by ω. Okay, so ω is a function, and what we are looking for is dω/dx. We will apply the chain rule:

dω/dx=dω/dt * dt/dx=d/dt(ω)*dt/dx

Now we simply put back dy/dx for ω to get

d2y/dx2=d/dt(dy/dx)*dt/dx as claimed. Hope that clears it up

5. Apr 1, 2012

### chiro

Only one note of caution with this approach: make sure dt/dx is defined which means dx/dt is non-zero (i.e. t changes with respect to x at the point you are considering and hence x also changes with respect to y)