derivative f(x)= 1/(-2+(3x+1)^2)

解

導関数

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

- \frac{18 x + 6}{( 9 x ^{2} + 6 x - 1 ) ^{2}}

解答ステップ

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

指数の規則を適用する:

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

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

連鎖法則を適用:

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

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

\frac{d}{d x} (- 2 + (3 x + 1)^{2}) = 18 x + 6

= - \frac{1}{( - 2 + ( 3 x + 1 ) ^{2} ) ^{2}} (18 x + 6)

簡素化

- \frac{1}{( - 2 + ( 3 x + 1 ) ^{2} ) ^{2}} (18 x + 6) : - \frac{18 x + 6}{( 9 x ^{2} + 6 x - 1 ) ^{2}}

= - \frac{18 x + 6}{( 9 x ^{2} + 6 x - 1 ) ^{2}}