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

Explanation

1. 01

   $\frac{dx}{dt} = e^{t}$ and $\frac{dy}{dt} = - e^{- t}$.

2. 02

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

3. 03

   At $t = \log 2$ the slope is $- \frac{1}{4}$.