derivative of x^{x+1}

解

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

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

解答ステップ

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

指数の規則を適用する:

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

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

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

連鎖法則を適用:

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

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

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

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

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

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