derivative f(x)=(x+2)^{x^2-2x+1}

解

導関数

f (x) = (x + 2)^{x^{2} - 2 x + 1}

(x^{2} - 2 x + 1 + 2 x^{2} ln (x + 2) + 2 x ln (x + 2) - 4 ln (x + 2)) (x + 2)^{x^{2} - 2 x}

解答ステップ

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

指数の規則を適用する:

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

(x + 2)^{x^{2} - 2 x + 1} = e^{(x^{2} - 2 x + 1) l n (x + 2)}

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

連鎖法則を適用:

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

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

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

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

簡素化

e^{(x^{2} - 2 x + 1) l n (x + 2)} ((2 x - 2) ln (x + 2) + \frac{1}{x + 2} (x^{2} - 2 x + 1)) : (x^{2} - 2 x + 1 + 2 x^{2} ln (x + 2) + 2 x ln (x + 2) - 4 ln (x + 2)) (x + 2)^{x^{2} - 2 x}

= (x^{2} - 2 x + 1 + 2 x^{2} ln (x + 2) + 2 x ln (x + 2) - 4 ln (x + 2)) (x + 2)^{x^{2} - 2 x}