Given $x=\cos t, y=\ln t^2$. Find the $\frac{d^2y}{dx^2}$ at $x=\frac{\pi}{2}$.
I'm confused about how we differentiate at $x=\frac{\pi}{2}$, because for this value we cannot find a $t$ value.
On the other hand, a similar problem I think:
For parametric equation $x=\sin^2t, y=cos^2t$, is it possible to find the derivative $\frac{dy}{dx}$ at $x=2$?
Although for $x=2$, there isn't any $t$, the parametric equation is equivalent to $y=1-x$. But which interval of $t$? Thank is advance.
Well, for your second question, since $y=1-x$, $\displaystyle \frac{dy}{dx}=-1$ always. But note, $x=2$ isn't even on the graph; perhaps you meant $t=2$.
Anyways, in parametric equations, $\displaystyle \frac{dy}{dx}$ is defined as $\displaystyle \frac{dy/dt}{dx/dt}$.
That will be an expression of $x$ and $y$. Then, $\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$.
(If you need to differentiate a $y$, just plug in the expression for $\displaystyle \frac{dy}{dx}$. )