(x^{1/x})^'

解

(x^{\frac{1}{x}})^{^{'}}

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

解答ステップ

(x^{\frac{1}{x}})^{'}

指数の規則を適用する:

a^{b} = e^{b l n (a)}

x^{\frac{1}{x}} = e^{\frac{1}{x} l n (x)}

= (e^{\frac{1}{x} l n (x)})^{^{'}}

連鎖法則を適用:

e^{\frac{1}{x} l n (x)} (\frac{1}{x} ln (x))^{'}

= e^{\frac{1}{x} l n (x)} (\frac{1}{x} ln (x))^{'}

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

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

簡素化

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

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