What is the derivative of ln(1+sin(2x)) ?

The derivative of ln(1+sin(2x)) is (2cos(2x))/(1+sin(2x))

What is the first derivative of ln(1+sin(2x)) ?

The first derivative of ln(1+sin(2x)) is (2cos(2x))/(1+sin(2x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

\frac{d}{d x} (ln (1 + sin (2 x)))

\frac{2 cos ( 2 x )}{1 + sin ( 2 x )}

Solution steps

\frac{d}{d x} (ln (1 + sin (2 x)))

Apply the chain rule:

\frac{1}{1 + sin ( 2 x )} \frac{d}{d x} (1 + sin (2 x))

= \frac{1}{1 + sin ( 2 x )} \frac{d}{d x} (1 + sin (2 x))

\frac{d}{d x} (1 + sin (2 x)) = 2 cos (2 x)

= \frac{1}{1 + sin ( 2 x )} \cdot 2 cos (2 x)

Simplify

\frac{1}{1 + sin ( 2 x )} \cdot 2 cos (2 x) : \frac{2 cos ( 2 x )}{1 + sin ( 2 x )}

= \frac{2 cos ( 2 x )}{1 + sin ( 2 x )}

What is the derivative of ln(1+sin(2x)) ?
The derivative of ln(1+sin(2x)) is (2cos(2x))/(1+sin(2x))
What is the first derivative of ln(1+sin(2x)) ?
The first derivative of ln(1+sin(2x)) is (2cos(2x))/(1+sin(2x))