# Parametric derivatives

1. Jan 6, 2014

### sooyong94

1. The problem statement, all variables and given/known data
Given a pair of parametric equations,
$x=f(t)$ and $y=g(t)$ ,
The first derivative is given by
$\frac{dy}{dx}=\frac{g'(t)}{f'(t)}$
and the second derivative is actually
$\frac{d}{dt}(\frac{dy}{dx})$
But why we cannot find the second derivative of a parametric equation by doing this?
$\frac{d^2 y}{dx^2}=\frac{g''(t}{f''(t)}$

2. Relevant equations

Parametric derivatives

3. The attempt at a solution
I know the parametric derivatives can be done by treating
$\frac{dy}{dt}$ and $\frac{dx}{dt}$ as a fraction, but why can't we do that to second derivatives?

2. Jan 6, 2014

### vanhees71

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{g'(t)}{f'(t)}.$$
Then you should get an idea, how to get the 2nd derivative.

3. Jan 6, 2014

### tiny-tim

hi sooyong94!
you mean $\frac{d}{dx}\left(\frac{dy}{dx}\right)$
no, that's dv/du where u(t) = f'(t), v(t) = g'(t)

4. Jan 6, 2014

### sooyong94

I don't get it though... I known I can use quotient rule to find the second derivative...

5. Jan 6, 2014

### Curious3141

Because you're not really treating them as fractions. You're using Chain Rule. World of difference.

6. Jan 6, 2014

### tiny-tim

yes, chain rule: dy/dx = dy/dt*dt/dx = dy/dt*(1/(dx/dt)) = g'(t)/f'(t) …

nothing to do with quotients

7. Jan 6, 2014

### sooyong94

I got it right now.... So it's about chain rule, not treating them as fractions.