What is the d/(dt)(a*e^{-9t}*cos(3t+b)) ?

The d/(dt)(a*e^{-9t}*cos(3t+b)) is a(-9e^{-9t}cos(3t+b)-3e^{-9t}sin(3t+b))

What is the first d/(dt)(a*e^{-9t}*cos(3t+b)) ?

The first d/(dt)(a*e^{-9t}*cos(3t+b)) is a(-9e^{-9t}cos(3t+b)-3e^{-9t}sin(3t+b))

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\frac{d}{d t} (a \cdot e^{- 9 t} \cdot cos (3 t + b))

a (- 9 e^{- 9 t} cos (3 t + b) - 3 e^{- 9 t} sin (3 t + b))

Solution steps

\frac{d}{d t} (a e^{- 9 t} cos (3 t + b))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= a \frac{d}{d t} (e^{- 9 t} cos (3 t + b))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{- 9 t}, g = cos (3 t + b)

= a (\frac{d}{d t} (e^{- 9 t}) cos (3 t + b) + \frac{d}{d t} (cos (3 t + b)) e^{- 9 t})

\frac{d}{d t} (e^{- 9 t}) = - 9 e^{- 9 t}

\frac{d}{d t} (cos (3 t + b)) = - sin (3 t + b) \cdot 3

= a ((- 9 e^{- 9 t}) cos (3 t + b) + (- sin (3 t + b) \cdot 3) e^{- 9 t})

Simplify

= a (- 9 e^{- 9 t} cos (3 t + b) - 3 e^{- 9 t} sin (3 t + b))

lo g_{10} (x^{2} + 1)

lo g_{10} (x^{2} + y^{2})

3^{x + 2}

3^{x + 1}

4 x ∣ x ∣

What is the d/(dt)(a*e^{-9t}*cos(3t+b)) ?
The d/(dt)(a*e^{-9t}*cos(3t+b)) is a(-9e^{-9t}cos(3t+b)-3e^{-9t}sin(3t+b))
What is the first d/(dt)(a*e^{-9t}*cos(3t+b)) ?
The first d/(dt)(a*e^{-9t}*cos(3t+b)) is a(-9e^{-9t}cos(3t+b)-3e^{-9t}sin(3t+b))