derivative of 1/(x^3+4x^2)

解

\frac{d}{d x} (\frac{1}{x ^{3} + 4 x ^{2}})

- \frac{3 x + 8}{x ^{3} ( x + 4 ) ^{2}}

解答ステップ

\frac{d}{d x} (\frac{1}{x ^{3} + 4 x ^{2}})

指数の規則を適用する:

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

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

連鎖法則を適用:

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

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

\frac{d}{d x} (x^{3} + 4 x^{2}) = 3 x^{2} + 8 x

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

簡素化

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

= - \frac{3 x + 8}{x ^{3} ( x + 4 ) ^{2}}