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))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

\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})

\frac{d}{d x} (csc^{7} (x))

\int \frac{x + 2}{( x ^{2} + 2 x + 10 ) ^{\frac{5}{2}}} d x

0 = 2 x + 1 - 2 y y^{^{'}}, y (- 2) = - 1

\frac{d}{d x} (sin (2 \cdot x))

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))