# What is the second derivative of f(t) = (3e^-2t) - (5e^-t) ?

Aug 5, 2018

$\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} = - \frac{5}{e} ^ t$

#### Explanation:

$f \left(t\right) = \left(3 {e}^{- 2} t\right) - \left(5 {e}^{- t}\right)$ or

$f \left(t\right) = \left(\frac{3}{e} ^ 2 t\right) - \left(5 {e}^{- t}\right)$ or

${f}^{'} \left(t\right) = \left(\frac{3}{e} ^ 2\right) - \left(5 {e}^{- t} \cdot \left(- 1\right)\right)$ or

${f}^{'} \left(t\right) = \left(\frac{3}{e} ^ 2\right) + \left(5 {e}^{- t}\right)$ or

$\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} = 0 + \left(5 {e}^{- t} \cdot \left(- 1\right)\right)$ or

$\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} = - \left(5 {e}^{- t}\right)$ or

$\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} = - \frac{5}{e} ^ t$ [Ans]