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

解

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

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

解答ステップ

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

指数の規則を適用する:

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

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

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

連鎖法則を適用:

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

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

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

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

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

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