derivative of 1/(2x+sin^3(x))

解

\frac{d}{d x} (\frac{1}{2 x + sin ^{3} ( x )})

- \frac{2 + 3 sin ^{2} ( x ) cos ( x )}{( 2 x + sin ^{3} ( x ) ) ^{2}}

解答ステップ

\frac{d}{d x} (\frac{1}{2 x + sin ^{3} ( x )})

指数の規則を適用する:

\frac{1}{a} = a^{- 1}

= \frac{d}{d x} ((2 x + sin^{3} (x))^{- 1})

連鎖法則を適用:

- \frac{1}{( 2 x + sin ^{3} ( x ) ) ^{2}} \frac{d}{d x} (2 x + sin^{3} (x))

= - \frac{1}{( 2 x + sin ^{3} ( x ) ) ^{2}} \frac{d}{d x} (2 x + sin^{3} (x))

\frac{d}{d x} (2 x + sin^{3} (x)) = 2 + 3 sin^{2} (x) cos (x)

= - \frac{1}{( 2 x + sin ^{3} ( x ) ) ^{2}} (2 + 3 sin^{2} (x) cos (x))

簡素化

- \frac{1}{( 2 x + sin ^{3} ( x ) ) ^{2}} (2 + 3 sin^{2} (x) cos (x)) : - \frac{2 + 3 sin ^{2} ( x ) cos ( x )}{( 2 x + sin ^{3} ( x ) ) ^{2}}

= - \frac{2 + 3 sin ^{2} ( x ) cos ( x )}{( 2 x + sin ^{3} ( x ) ) ^{2}}