derivative of 1/(2xsqrt(ln(x+1)))

解

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

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

解答ステップ

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

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

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

指数の規則を適用する:

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

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

連鎖法則を適用:

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

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

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

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

簡素化

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

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