What is the d/(dt)(-1+cos^2(t)) ?

The d/(dt)(-1+cos^2(t)) is -sin(2t)

What is the first d/(dt)(-1+cos^2(t)) ?

The first d/(dt)(-1+cos^2(t)) is -sin(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

- sin (2 t)

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

Apply the chain rule:

2 sin (t) \frac{d}{d t} (sin (t))

= 2 sin (t) \frac{d}{d t} (sin (t))

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

= - 2 sin (t) cos (t)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

= - sin (2 t)

What is the d/(dt)(-1+cos^2(t)) ?
The d/(dt)(-1+cos^2(t)) is -sin(2t)
What is the first d/(dt)(-1+cos^2(t)) ?
The first d/(dt)(-1+cos^2(t)) is -sin(2t)