derivative of ln(1+sin(2x))

解

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

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

解答ステップ

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

連鎖法則を適用:

\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)

簡素化

\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 )}