What is the d/(dt)((sin(t))/(1-cos(t))) ?

The d/(dt)((sin(t))/(1-cos(t))) is 1/(-1+cos(t))

What is the first d/(dt)((sin(t))/(1-cos(t))) ?

The first d/(dt)((sin(t))/(1-cos(t))) is 1/(-1+cos(t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d t} (\frac{sin ( t )}{1 - cos ( t )})

\frac{1}{- 1 + cos ( t )}

Solution steps

\frac{d}{d t} (\frac{sin ( t )}{1 - cos ( t )})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d t} ( sin ( t ) ) ( 1 - cos ( t ) ) - \frac{d}{d t} ( 1 - cos ( t ) ) sin ( t )}{( 1 - cos ( t ) ) ^{2}}

\frac{d}{d t} (sin (t)) = cos (t)

\frac{d}{d t} (1 - cos (t)) = sin (t)

= \frac{cos ( t ) ( 1 - cos ( t ) ) - sin ( t ) sin ( t )}{( 1 - cos ( t ) ) ^{2}}

Simplify

\frac{cos ( t ) ( 1 - cos ( t ) ) - sin ( t ) sin ( t )}{( 1 - cos ( t ) ) ^{2}} : \frac{1}{- 1 + cos ( t )}

= \frac{1}{- 1 + cos ( t )}

d x + \frac{1}{y ^{4}} d y = 0

\int_{1}^{3} 3 x^{2} + 4 x + 3 d x

\int_{0}^{2} (y - 2) (2 y + 1) d y

\int 1 + cos (π x) d x

\frac{\partial}{\partial y} (cos (x) sin (y))

What is the d/(dt)((sin(t))/(1-cos(t))) ?
The d/(dt)((sin(t))/(1-cos(t))) is 1/(-1+cos(t))
What is the first d/(dt)((sin(t))/(1-cos(t))) ?
The first d/(dt)((sin(t))/(1-cos(t))) is 1/(-1+cos(t))