A curve is given by $x = 3t^{2} + 1$, $y = t^{3} - 2t$. What is its slope $\frac{dy}{dx}$ at $t = 1$?

Explanation

1. 01

   $\frac{dx}{dt} = 6t = 6$ and $\frac{dy}{dt} = 3t^{2} - 2 = 1$ at $t = 1$.

2. 02

   $\frac{dy}{dx} = \frac{1}{6}$.