What is the d/(dt)((x)(xsin(wt+c))) ?

The d/(dt)((x)(xsin(wt+c))) is x^2cos(wt+c)w

What is the first d/(dt)((x)(xsin(wt+c))) ?

The first d/(dt)((x)(xsin(wt+c))) is x^2cos(wt+c)w

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d t} ((x) (x sin (wt + c)))

x^{2} cos (wt + c) w

Solution steps

\frac{d}{d t} ((x) (x sin (wt + c)))

Simplify

(x) (x sin (wt + c)) : x^{2} sin (wt + c)

= \frac{d}{d t} (x^{2} sin (wt + c))

Treat

c, x, w

as constants

Take the constant out:

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

= x^{2} \frac{d}{d t} (sin (wt + c))

Apply the chain rule:

cos (wt + c) \frac{d}{d t} (wt + c)

= cos (wt + c) \frac{d}{d t} (wt + c)

\frac{d}{d t} (wt + c) = w

= x^{2} cos (wt + c) w

15 a^{2} b + 3 a t + 10 abm + 2 m t

4 x^{2} - 12 x^{2}

3 x^{3} + 3 x^{2} - 2 x^{0.3}

f (x) = 7 x + 4

(4 x^{2} + 2 y) + (3 x + y^{2})

What is the d/(dt)((x)(xsin(wt+c))) ?
The d/(dt)((x)(xsin(wt+c))) is x^2cos(wt+c)w
What is the first d/(dt)((x)(xsin(wt+c))) ?
The first d/(dt)((x)(xsin(wt+c))) is x^2cos(wt+c)w