derivative g(x)=(1-x^2)^{-x}

解

導関数

g (x) = (1 - x^{2})^{- x}

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

解答ステップ

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

指数の規則を適用する:

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

(1 - x^{2})^{- x} = e^{(- x) l n (1 - x^{2})}

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

連鎖法則を適用:

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

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

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

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

簡素化

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

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