What is the d/(dt)(A*(cos(ωt+φ))) ?

The d/(dt)(A*(cos(ωt+φ))) is -Aωsin(ωt+φ)

What is the first d/(dt)(A*(cos(ωt+φ))) ?

The first d/(dt)(A*(cos(ωt+φ))) is -Aωsin(ωt+φ)

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\frac{d}{d t} (A \cdot (cos (ω t + φ)))

- A ω sin (ω t + φ)

Solution steps

\frac{d}{d t} (A (cos (ω t + φ)))

Treat

A, ω, φ

as a constant

Take the constant out:

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

= A \frac{d}{d t} (cos (ω t + φ))

Apply the chain rule:

- sin (ω t + φ) \frac{d}{d t} (ω t + φ)

= - sin (ω t + φ) \frac{d}{d t} (ω t + φ)

\frac{d}{d t} (ω t + φ) = ω

= A (- sin (ω t + φ) ω)

Simplify

= - A ω sin (ω t + φ)

y^{^{'}} (x) = \frac{x ^{2}}{1 - y ^{2}}, y (4) = 2

\int 2 x e^{x} d x

y = (x^{2} + 1)^{3}

\int \frac{1 + cos ( 2 x )}{sin ^{2} ( 2 x )} d x

\frac{\partial}{\partial y} (x + 3 y)

What is the d/(dt)(A*(cos(ωt+φ))) ?
The d/(dt)(A*(cos(ωt+φ))) is -Aωsin(ωt+φ)
What is the first d/(dt)(A*(cos(ωt+φ))) ?
The first d/(dt)(A*(cos(ωt+φ))) is -Aωsin(ωt+φ)