Find the derivative $dy/dx$ from the parametric equations for $x$ and $y$ 
Let 
  \begin{cases}
y=2t^2-t+1 \\
x=\sin(t)
\end{cases}
find $\frac{dy}{dx}$

Is this all that I need to do? $$\frac{4t-1}{\cos(t)}$$

Just use the chain rule. Take the derivative of the first with respect to x:
$\frac{dy}{dx}=4t\frac{dt}{dx}-\frac{dt}{dx}$
But now we need to know what $\frac{dt}{dx}$ is. Take the derivative of the second with respect to $x$ to get:
$\frac{dx}{dx}=\cos(t)\frac{dt}{dx}$, or $\frac{dt}{dx}=\frac{1}{\cos(t)}$.
Indeed, we have $\frac{dy}{dx}=4t\frac{1}{\cos(t)}-\frac{1}{\cos(t)}=\frac{4t-1}{\cos(t)}$.
