derivative of 2x^{sqrt(x)}

解

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

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

解答ステップ

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

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d x} (x^{x})

指数の規則を適用する:

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

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

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

連鎖法則を適用:

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

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

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

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

簡素化

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

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