derivative of x^{x^x}

解

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

x^{x^{x}} (x^{x} ln (x) (ln (x) + 1) + x^{x - 1})

解答ステップ

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

指数の規則を適用する:

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

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

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

連鎖法則を適用:

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

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

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

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

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

= x^{x^{x}} (x^{x} ln (x) (ln (x) + 1) + x^{x - 1})