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

解

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

- \frac{9 x + 2}{x ^{3} ( 3 x + 1 ) ^{2}}

解答ステップ

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

指数の規則を適用する:

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

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

連鎖法則を適用:

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

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

\frac{d}{d x} (x^{2} (3 x + 1)) = 9 x^{2} + 2 x

= - \frac{1}{( x ^{2} ( 3 x + 1 ) ) ^{2}} (9 x^{2} + 2 x)

簡素化

- \frac{1}{( x ^{2} ( 3 x + 1 ) ) ^{2}} (9 x^{2} + 2 x) : - \frac{9 x + 2}{x ^{3} ( 3 x + 1 ) ^{2}}

= - \frac{9 x + 2}{x ^{3} ( 3 x + 1 ) ^{2}}