derivative of 1/(sqrt(1+4x^2+9x^4))

解

\frac{d}{d x} (\frac{1}{1 + 4 x ^{2} + 9 x ^{4}})

- \frac{2 x ( 9 x ^{2} + 2 )}{( 1 + 4 x ^{2} + 9 x ^{4} ) ^{\frac{3}{2}}}

解答ステップ

\frac{d}{d x} (\frac{1}{1 + 4 x ^{2} + 9 x ^{4}})

指数の規則を適用する:

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

= \frac{d}{d x} ((1 + 4 x^{2} + 9 x^{4})^{- \frac{1}{2}})

連鎖法則を適用:

- \frac{1}{2 ( 1 + 4 x ^{2} + 9 x ^{4} ) ^{\frac{3}{2}}} \frac{d}{d x} (1 + 4 x^{2} + 9 x^{4})

= - \frac{1}{2 ( 1 + 4 x ^{2} + 9 x ^{4} ) ^{\frac{3}{2}}} \frac{d}{d x} (1 + 4 x^{2} + 9 x^{4})

\frac{d}{d x} (1 + 4 x^{2} + 9 x^{4}) = 36 x^{3} + 8 x

= - \frac{1}{2 ( 1 + 4 x ^{2} + 9 x ^{4} ) ^{\frac{3}{2}}} (36 x^{3} + 8 x)

簡素化

- \frac{1}{2 ( 1 + 4 x ^{2} + 9 x ^{4} ) ^{\frac{3}{2}}} (36 x^{3} + 8 x) : - \frac{2 x ( 9 x ^{2} + 2 )}{( 1 + 4 x ^{2} + 9 x ^{4} ) ^{\frac{3}{2}}}

= - \frac{2 x ( 9 x ^{2} + 2 )}{( 1 + 4 x ^{2} + 9 x ^{4} ) ^{\frac{3}{2}}}