What is the d/(dt)(ae^{-t}+se^{-4t}) ?

The d/(dt)(ae^{-t}+se^{-4t}) is -ae^{-t}-4se^{-4t}

What is the first d/(dt)(ae^{-t}+se^{-4t}) ?

The first d/(dt)(ae^{-t}+se^{-4t}) is -ae^{-t}-4se^{-4t}

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\frac{d}{d t} (a e^{- t} + s e^{- 4 t})

- a e^{- t} - 4 s e^{- 4 t}

Solution steps

\frac{d}{d t} (a e^{- t} + s e^{- 4 t})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (a e^{- t}) + \frac{d}{d t} (s e^{- 4 t})

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

\frac{d}{d t} (s e^{- 4 t}) = - 4 s e^{- 4 t}

= - a e^{- t} - 4 s e^{- 4 t}

lo g_{10} (2) (x^{58})

A u (O H) 3 + H^{2} G e O^{4}

1 - 2 x y - (x^{2} + y^{2})

e^{4 x} \frac{d y}{d x} + e^{4 x} 4 y

(- 1) - (ln (2))

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