What is the first (sin(2x))^' ?

The first (sin(2x))^' is cos(2x)*2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(sin(2x))^'

(sin (2 x))^{^{'}}

cos (2 x) \cdot 2

Solution steps

(sin (2 x))^{'}

Apply the chain rule:

cos (2 x) (2 x)^{'}

= cos (2 x) (2 x)^{'}

(2 x)^{'} = 2

= cos (2 x) \cdot 2

\frac{\partial}{\partial x} (x^{2} + 6 x y + y^{6})

\int \frac{8}{( 4 x + 5 ) ^{3}} d x

\frac{1}{1 + sin ( x )}

\frac{1}{( 1 - 0.8 \cdot cos ^{2} ( x ) )}

\frac{5}{( s ^{2} + 5 )}