derivative of 2x^{4+7x+5}

解

\frac{d}{d x} (2 x^{4 + 7 x + 5})

2 x^{7 x + 9} (7 ln (x) + \frac{7 x + 9}{x})

解答ステップ

\frac{d}{d x} (2 x^{4 + 7 x + 5})

簡素化

2 x^{4 + 7 x + 5} : 2 x^{7 x + 9}

= \frac{d}{d x} (2 x^{7 x + 9})

定数を除去:

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

= 2 \frac{d}{d x} (x^{7 x + 9})

指数の規則を適用する:

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

x^{7 x + 9} = e^{(7 x + 9) l n (x)}

= 2 \frac{d}{d x} (e^{(7 x + 9) l n (x)})

連鎖法則を適用:

e^{(7 x + 9) l n (x)} \frac{d}{d x} ((7 x + 9) ln (x))

= e^{(7 x + 9) l n (x)} \frac{d}{d x} ((7 x + 9) ln (x))

\frac{d}{d x} ((7 x + 9) ln (x)) = 7 ln (x) + \frac{7 x + 9}{x}

= 2 e^{(7 x + 9) l n (x)} (7 ln (x) + \frac{7 x + 9}{x})

e^{(7 x + 9) l n (x)} = x^{7 x + 9}

= 2 x^{7 x + 9} (7 ln (x) + \frac{7 x + 9}{x})