What is the d/(dt)(5e^{-2t}+3te^{-2t}-2) ?

The d/(dt)(5e^{-2t}+3te^{-2t}-2) is -6e^{-2t}t-7e^{-2t}

What is the first d/(dt)(5e^{-2t}+3te^{-2t}-2) ?

The first d/(dt)(5e^{-2t}+3te^{-2t}-2) is -6e^{-2t}t-7e^{-2t}

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\frac{d}{d t} (5 e^{- 2 t} + 3 t e^{- 2 t} - 2)

- 6 e^{- 2 t} t - 7 e^{- 2 t}

Solution steps

\frac{d}{d t} (5 e^{- 2 t} + 3 t e^{- 2 t} - 2)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (5 e^{- 2 t}) + \frac{d}{d t} (3 t e^{- 2 t}) - \frac{d}{d t} (2)

\frac{d}{d t} (5 e^{- 2 t}) = - 10 e^{- 2 t}

\frac{d}{d t} (3 t e^{- 2 t}) = 3 (e^{- 2 t} - 2 e^{- 2 t} t)

\frac{d}{d t} (2) = 0

= - 10 e^{- 2 t} + 3 (e^{- 2 t} - 2 e^{- 2 t} t) - 0

Simplify

- 10 e^{- 2 t} + 3 (e^{- 2 t} - 2 e^{- 2 t} t) - 0 : - 6 e^{- 2 t} t - 7 e^{- 2 t}

= - 6 e^{- 2 t} t - 7 e^{- 2 t}

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (1) = 0, y^{^{'}} (1) = 4

x cos (x)

\int \frac{2 + x}{4 x ^{2} - 4 x - 3} d x

f (x) = ln (x - 1)

\int (12 x^{2} - 8 x + 2) d x

What is the d/(dt)(5e^{-2t}+3te^{-2t}-2) ?
The d/(dt)(5e^{-2t}+3te^{-2t}-2) is -6e^{-2t}t-7e^{-2t}
What is the first d/(dt)(5e^{-2t}+3te^{-2t}-2) ?
The first d/(dt)(5e^{-2t}+3te^{-2t}-2) is -6e^{-2t}t-7e^{-2t}