derivative g(x)=2^{x^2+1x}

解

導関数

g (x) = 2^{x^{2} + 1 x}

ln (2) \cdot 2^{x^{2} + x} (2 x + 1)

解答ステップ

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

簡素化

2^{x^{2} + 1 \cdot x} : 2^{x^{2} + x}

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

指数の規則を適用する:

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

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

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

連鎖法則を適用:

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

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

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

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

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

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