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? $$\frac{4t-1}{\cos(t)}$$
 A: HINT: we have $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ and $$\frac{dy}{dt}=4t-1$$ and $$\frac{dx}{dt}=\cos(t)$$ from here you will get the answer
A: As others have said,
$$
\frac{dy}{dt} = 4t-1 \quad\mbox{and}\quad \frac{dx}{dt} = \cos t \implies \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t-1}{\cos t}.
$$
But now we must eliminate the parameter $t$, because we'd like to have $y$ as a function of $x$ in writing $dy/dx$. 
For the numerator of $dy/dx$, we use
$$
x = \sin t \implies t = \arcsin x.
$$
For the denominator of $dy/dx$, we square $x = \cos t$ to get
$$
x^2 = \sin^2 t = 1 - \cos^2 t \implies \cos^2 t = 1 - x^2 \implies \cos t = \pm \sqrt{1 - x^2}.
$$
To get rid of the $\pm$, you should specify the range of allowed values for $t$, and hence for $x$ to make $\arcsin t$. For example, restricting $t \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ results in $x \in [-1,1]$ and we should take $\cos t = + \sqrt{1 - x^2}$, yielding
$$
\frac{dy}{dx} = \frac{4 \arcsin(x) - 1}{\sqrt{1 - x^2}}.
$$
A: 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 $\frac{dy}{dx}=4t\frac{1}{\cos(t)}-\frac{1}{\cos(t)}=\frac{4t-1}{\cos(t)}$.
