derivative of 1/(sqrt(2x+1))

解

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

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

解答ステップ

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

指数の規則を適用する:

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

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

連鎖法則を適用:

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

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

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

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

簡素化

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

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