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

The d/(dt)(2e^{-t}t-e^{-t}t^2) is e^{-t}t^2-4e^{-t}t+2e^{-t}

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

The first d/(dt)(2e^{-t}t-e^{-t}t^2) is e^{-t}t^2-4e^{-t}t+2e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

e^{- t} t^{2} - 4 e^{- t} t + 2 e^{- t}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 2 (- e^{- t} t + e^{- t}) - (- e^{- t} t^{2} + 2 e^{- t} t)

Simplify

2 (- e^{- t} t + e^{- t}) - (- e^{- t} t^{2} + 2 e^{- t} t) : e^{- t} t^{2} - 4 e^{- t} t + 2 e^{- t}

= e^{- t} t^{2} - 4 e^{- t} t + 2 e^{- t}

f (x) = ln (2 x^{5} (5 x - 1)^{3})

y = 3 x + 1

\frac{d}{d x} ((9 x^{6} + 4 x^{3})^{4})

t \to π lim (sin (\frac{3}{4} t))

(ln (x))^{5 c o s (x)}

What is the d/(dt)(2e^{-t}t-e^{-t}t^2) ?
The d/(dt)(2e^{-t}t-e^{-t}t^2) is e^{-t}t^2-4e^{-t}t+2e^{-t}
What is the first d/(dt)(2e^{-t}t-e^{-t}t^2) ?
The first d/(dt)(2e^{-t}t-e^{-t}t^2) is e^{-t}t^2-4e^{-t}t+2e^{-t}