derivative f(x)=(x^2+3)^3(x^3+3)^2

解

導関数

f (x) = (x^{2} + 3)^{3} (x^{3} + 3)^{2}

6 x (x^{2} + 3)^{2} (x^{3} + 3)^{2} + 6 x^{2} (x^{3} + 3) (x^{2} + 3)^{3}

解答ステップ

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

積の法則を適用:

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

f = (x^{2} + 3)^{3}, g = (x^{3} + 3)^{2}

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

\frac{d}{d x} ((x^{2} + 3)^{3}) = 6 x (x^{2} + 3)^{2}

\frac{d}{d x} ((x^{3} + 3)^{2}) = 6 x^{2} (x^{3} + 3)

= 6 x (x^{2} + 3)^{2} (x^{3} + 3)^{2} + 6 x^{2} (x^{3} + 3) (x^{2} + 3)^{3}