derivative of x^{-1}ln(x)

解

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

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

解答ステップ

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

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{- 1}, g = ln (x)

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

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

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

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

簡素化

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

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