derivative of 3^{sqrt(x^3-x+1)}

解

\frac{d}{d x} (3^{x^{3} - x + 1})

\frac{ln ( 3 ) \cdot 3 ^{x^{3} - x + 1} ( 3 x ^{2} - 1 )}{2 x ^{3} - x + 1}

解答ステップ

\frac{d}{d x} (3^{x^{3} - x + 1})

指数の規則を適用する:

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

3^{x^{3} - x + 1} = e^{x^{3} - x + 1 l n (3)}

= \frac{d}{d x} (e^{x^{3} - x + 1 l n (3)})

連鎖法則を適用:

e^{x^{3} - x + 1 l n (3)} \frac{d}{d x} (x^{3} - x + 1 ln (3))

= e^{x^{3} - x + 1 l n (3)} \frac{d}{d x} (x^{3} - x + 1 ln (3))

\frac{d}{d x} (x^{3} - x + 1 ln (3)) = \frac{ln ( 3 ) ( 3 x ^{2} - 1 )}{2 x ^{3} - x + 1}

= e^{x^{3} - x + 1 l n (3)} \frac{ln ( 3 ) ( 3 x ^{2} - 1 )}{2 x ^{3} - x + 1}

簡素化

e^{x^{3} - x + 1 l n (3)} \frac{ln ( 3 ) ( 3 x ^{2} - 1 )}{2 x ^{3} - x + 1} : \frac{ln ( 3 ) \cdot 3 ^{x^{3} - x + 1} ( 3 x ^{2} - 1 )}{2 x ^{3} - x + 1}

= \frac{ln ( 3 ) \cdot 3 ^{x^{3} - x + 1} ( 3 x ^{2} - 1 )}{2 x ^{3} - x + 1}