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

Explanation

1. 01

   $\frac{dx}{dt} = \frac{1}{t}$ and $\frac{dy}{dt} = 2 t$.

2. 02

   $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = 2 t^{2}$.

3. 03

   At $t = 1$ the slope is $2$.