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

解

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

- \frac{3 ( 5 x + 1 )}{( x + 1 ) ^{4} ( 3 x - 1 ) ^{3}}

解答ステップ

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

指数の規則を適用する:

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

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

連鎖法則を適用:

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

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

\frac{d}{d x} ((x + 1)^{3} (3 x - 1)^{2}) = 3 (x + 1)^{2} (3 x - 1)^{2} + 6 (3 x - 1) (x + 1)^{3}

= - \frac{1}{( ( x + 1 ) ^{3} ( 3 x - 1 ) ^{2} ) ^{2}} (3 (x + 1)^{2} (3 x - 1)^{2} + 6 (3 x - 1) (x + 1)^{3})

簡素化

- \frac{1}{( ( x + 1 ) ^{3} ( 3 x - 1 ) ^{2} ) ^{2}} (3 (x + 1)^{2} (3 x - 1)^{2} + 6 (3 x - 1) (x + 1)^{3}) : - \frac{3 ( 5 x + 1 )}{( x + 1 ) ^{4} ( 3 x - 1 ) ^{3}}

= - \frac{3 ( 5 x + 1 )}{( x + 1 ) ^{4} ( 3 x - 1 ) ^{3}}