derivative R(x)=[x^2(x^2+6x)]^4

解答

导数

R (x) = [x^{2} (x^{2} + 6 x)]^{4}

8 x^{7} (x^{2} + 6 x)^{4} + 4 (x^{2} + 6 x)^{3} (2 x + 6) x^{8}

求解步骤

\frac{d}{d x} ((x^{2} (x^{2} + 6 x))^{4})

化简

(x^{2} (x^{2} + 6 x))^{4} : x^{8} (x^{2} + 6 x)^{4}

= \frac{d}{d x} (x^{8} (x^{2} + 6 x)^{4})

使用微分乘法定则:

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

f = x^{8}, g = (x^{2} + 6 x)^{4}

= \frac{d}{d x} (x^{8}) (x^{2} + 6 x)^{4} + \frac{d}{d x} ((x^{2} + 6 x)^{4}) x^{8}

\frac{d}{d x} (x^{8}) = 8 x^{7}

\frac{d}{d x} ((x^{2} + 6 x)^{4}) = 4 (x^{2} + 6 x)^{3} (2 x + 6)

= 8 x^{7} (x^{2} + 6 x)^{4} + 4 (x^{2} + 6 x)^{3} (2 x + 6) x^{8}

a re a y = \frac{1}{16} (e^{8 x} + e^{- 8 x}), - 4 \leq x \leq 4

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x ^{2} - 1})

\int t cos (8 t) d t

\int - y + 1 d y

\frac{d}{d x} (5 arcsin (3) x^{3})