derivative (x+1)^x

解

導関数

(x + 1)^{x}

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

解答ステップ

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

指数の規則を適用する:

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

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

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

連鎖法則を適用:

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

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

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

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

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

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