derivative of (x^2+1^{2x-x^2})

解

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

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

解答ステップ

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

指数の規則を適用する:

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

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

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

連鎖法則を適用:

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

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

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

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

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

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