# How do you find the derivative of y = [e^(-1) + e^(t)]^3?

The answer is : $3 {\left({e}^{- 1} + {e}^{t}\right)}^{2} \cdot {e}^{t}$
$\dot{y} = \dot{{\left({e}^{- 1} + {e}^{t}\right)}^{3}} = 3 {\left({e}^{- 1} + {e}^{t}\right)}^{2} \cdot \dot{\left({e}^{- 1} + {e}^{t}\right)}$
$= 3 {\left({e}^{- 1} + {e}^{t}\right)}^{2} \cdot {e}^{t}$