# I Second derivative of a curve defined by parametric equations

1. Mar 1, 2017

### Mr Davis 97

Quick question. I know that if we have a curve defined by $x=f(t)$ and $y=g(t)$, then the slope of the tangent line is $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. I am trying to find the second derivative, which would be $\displaystyle \frac{d}{dx}\frac{dy}{dx} = \frac{\frac{d}{dx}\frac{dy}{dt}}{\frac{dx}{dt}}$. Now, the final correct formula is $\displaystyle \frac{d^2 y}{dx^2} = \frac{\frac{d}{dt}\frac{dy}{dx}}{\frac{dx}{dt}}$. My question is, why is it valid that $\displaystyle \frac{d}{dx} \frac{dy}{dt} = \frac{d}{dt} \frac{dy}{dx}$?

2. Mar 1, 2017

### andrewkirk

The formula can be calculated by the chain rule, treating $x$ as the intermediate function (of $t$), as:
$$\frac{d}{dx}\left(\frac{dy}{dx}\right)= \left(\frac d{dt}\frac{dy}{dx}\right)\frac{dt}{dx}$$
And that is equal to
$$\left(\frac d{dt}\frac{dy}{dx}\right)\ /\ \frac{dx}{dt}$$
which is the same as what you describe as the final correct formula.

How did you get your own formula? It doesn't look correct to me. Doing it your way I get:
$$\frac{d}{dx}\left(\frac{dy}{dx}\right)= \frac{d}{dx}\left(\frac{dy}{dt}\ /\ \frac{dx}{dt}\right) =\left[\frac{dx}{dt} \frac{d}{dx}\frac{dy}{dt}- \frac{dy}{dt} \frac{d}{dx}\frac{dx}{dt} \right]\ /\ \left(\frac{dx}{dt}\right)^2$$
(using the quotient rule for the second step), which is not like what you wrote.

3. Mar 1, 2017

### Mr Davis 97

Okay, I see how you do it now, using the chain rule. However, why is it not correct reasoning that if $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, then $\displaystyle \frac{d}{dx}\frac{dy}{dx} = \frac{d}{dx}\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$?

4. Mar 1, 2017

### andrewkirk

It is correct. But what you wrote here is different from what you wrote in the OP. Perhaps your LaTeX got a bit muddled back there?