derivative y=(2x-9)^4(x^2+x+1)^5

解

導関数

y = (2 x - 9)^{4} (x^{2} + x + 1)^{5}

8 (2 x - 9)^{3} (x^{2} + x + 1)^{5} + 5 (x^{2} + x + 1)^{4} (2 x + 1) (2 x - 9)^{4}

解答ステップ

\frac{d}{d x} ((2 x - 9)^{4} (x^{2} + x + 1)^{5})

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = (2 x - 9)^{4}, g = (x^{2} + x + 1)^{5}

= \frac{d}{d x} ((2 x - 9)^{4}) (x^{2} + x + 1)^{5} + \frac{d}{d x} ((x^{2} + x + 1)^{5}) (2 x - 9)^{4}

\frac{d}{d x} ((2 x - 9)^{4}) = 8 (2 x - 9)^{3}

\frac{d}{d x} ((x^{2} + x + 1)^{5}) = 5 (x^{2} + x + 1)^{4} (2 x + 1)

= 8 (2 x - 9)^{3} (x^{2} + x + 1)^{5} + 5 (x^{2} + x + 1)^{4} (2 x + 1) (2 x - 9)^{4}