What is the d/(dt)((4cos(t)-2)e^{it}) ?

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

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

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

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\frac{d}{d t} ((4 cos (t) - 2) e^{i t})

(- 4 sin (2 t) + 2 sin (t)) + i (- 2 cos (t) + 4 cos (2 t))

Solution steps

\frac{d}{d t} ((4 cos (t) - 2) e^{i t})

Simplify

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

= \frac{d}{d t} ((4 cos^{2} (t) - 2 cos (t)) + i (- 2 sin (t) + 4 cos (t) sin (t)))

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (4 cos^{2} (t) - 2 cos (t)) + \frac{d}{d t} (i (- 2 sin (t) + 4 cos (t) sin (t)))

\frac{d}{d t} (4 cos^{2} (t) - 2 cos (t)) = - 4 sin (2 t) + 2 sin (t)

\frac{d}{d t} (i (- 2 sin (t) + 4 cos (t) sin (t))) = i (- 2 cos (t) + 4 cos (2 t))

= - 4 sin (2 t) + 2 sin (t) + i (- 2 cos (t) + 4 cos (2 t))

Rewrite

- 4 sin (2 t) + 2 sin (t) + i (- 2 cos (t) + 4 cos (2 t))

in standard complex form:

(- 4 sin (2 t) + 2 sin (t)) + (- 2 cos (t) + 4 cos (2 t)) i

= (- 4 sin (2 t) + 2 sin (t)) + i (- 2 cos (t) + 4 cos (2 t))

\frac{d}{d x} ((ln (4 x))^{2})

y^{^{'}} = 5 y + y^{3}

\int_{- π}^{π} x^{2} cos (n x) d x

\int_{0}^{4} π (x)^{2} d x

y^{^{'}} + 3 y = 15 - e^{t}

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