derivative log_{e}(sin(2x)+cot(2x))

解

導関数

lo g_{e} (sin (2 x) + cot (2 x))

\frac{2 cos ( 2 x ) - 2 csc ^{2} ( 2 x )}{sin ( 2 x ) + cot ( 2 x )}

解答ステップ

\frac{d}{d x} (lo g_{e} (sin (2 x) + cot (2 x)))

簡素化

lo g_{e} (sin (2 x) + cot (2 x)) : ln (sin (2 x) + cot (2 x))

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

連鎖法則を適用:

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

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

\frac{d}{d x} (sin (2 x) + cot (2 x)) = cos (2 x) \cdot 2 - csc^{2} (2 x) \cdot 2

= \frac{1}{sin ( 2 x ) + cot ( 2 x )} (cos (2 x) \cdot 2 - csc^{2} (2 x) \cdot 2)

簡素化

\frac{1}{sin ( 2 x ) + cot ( 2 x )} (cos (2 x) \cdot 2 - csc^{2} (2 x) \cdot 2) : \frac{2 cos ( 2 x ) - 2 csc ^{2} ( 2 x )}{sin ( 2 x ) + cot ( 2 x )}

= \frac{2 cos ( 2 x ) - 2 csc ^{2} ( 2 x )}{sin ( 2 x ) + cot ( 2 x )}