derivative of 1/(4x^3+5x^2-7x+8)

解

\frac{d}{d x} (\frac{1}{4 x ^{3} + 5 x ^{2} - 7 x + 8})

- \frac{12 x ^{2} + 10 x - 7}{( 4 x ^{3} + 5 x ^{2} - 7 x + 8 ) ^{2}}

解答ステップ

\frac{d}{d x} (\frac{1}{4 x ^{3} + 5 x ^{2} - 7 x + 8})

指数の規則を適用する:

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

= \frac{d}{d x} ((4 x^{3} + 5 x^{2} - 7 x + 8)^{- 1})

連鎖法則を適用:

- \frac{1}{( 4 x ^{3} + 5 x ^{2} - 7 x + 8 ) ^{2}} \frac{d}{d x} (4 x^{3} + 5 x^{2} - 7 x + 8)

= - \frac{1}{( 4 x ^{3} + 5 x ^{2} - 7 x + 8 ) ^{2}} \frac{d}{d x} (4 x^{3} + 5 x^{2} - 7 x + 8)

\frac{d}{d x} (4 x^{3} + 5 x^{2} - 7 x + 8) = 12 x^{2} + 10 x - 7

= - \frac{1}{( 4 x ^{3} + 5 x ^{2} - 7 x + 8 ) ^{2}} (12 x^{2} + 10 x - 7)

簡素化

- \frac{1}{( 4 x ^{3} + 5 x ^{2} - 7 x + 8 ) ^{2}} (12 x^{2} + 10 x - 7) : - \frac{12 x ^{2} + 10 x - 7}{( 4 x ^{3} + 5 x ^{2} - 7 x + 8 ) ^{2}}

= - \frac{12 x ^{2} + 10 x - 7}{( 4 x ^{3} + 5 x ^{2} - 7 x + 8 ) ^{2}}