derivative 2^{(x^2-3x+8)}

解

導関数

2^{(x^{2} - 3 x + 8)}

ln (2) \cdot 2^{x^{2} - 3 x + 8} (2 x - 3)

解答ステップ

\frac{d}{d x} (2^{x^{2} - 3 x + 8})

指数の規則を適用する:

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

2^{x^{2} - 3 x + 8} = e^{(x^{2} - 3 x + 8) l n (2)}

= \frac{d}{d x} (e^{(x^{2} - 3 x + 8) l n (2)})

連鎖法則を適用:

e^{(x^{2} - 3 x + 8) l n (2)} \frac{d}{d x} ((x^{2} - 3 x + 8) ln (2))

= e^{(x^{2} - 3 x + 8) l n (2)} \frac{d}{d x} ((x^{2} - 3 x + 8) ln (2))

\frac{d}{d x} ((x^{2} - 3 x + 8) ln (2)) = ln (2) (2 x - 3)

= e^{(x^{2} - 3 x + 8) l n (2)} ln (2) (2 x - 3)

e^{(x^{2} - 3 x + 8) l n (2)} = 2^{x^{2} - 3 x + 8}

= ln (2) \cdot 2^{x^{2} - 3 x + 8} (2 x - 3)