derivative h(x)=(x-4)(x/2+1)^3

解答

导数

h (x) = (x - 4) (\frac{x}{2} + 1)^{3}

\frac{2 x ^{3} + 3 x ^{2} - 12 x - 20}{4}

求解步骤

\frac{d}{d x} ((x - 4) (\frac{x}{2} + 1)^{3})

化简

(x - 4) (\frac{x}{2} + 1)^{3} : \frac{x ^{4} + 2 x ^{3} - 12 x ^{2} - 40 x}{8} - 4

= \frac{d}{d x} (\frac{x ^{4} + 2 x ^{3} - 12 x ^{2} - 40 x}{8} - 4)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{x ^{4} + 2 x ^{3} - 12 x ^{2} - 40 x}{8}) - \frac{d}{d x} (4)

\frac{d}{d x} (\frac{x ^{4} + 2 x ^{3} - 12 x ^{2} - 40 x}{8}) = \frac{2 x ^{3} + 3 x ^{2} - 12 x - 20}{4}

\frac{d}{d x} (4) = 0

= \frac{2 x ^{3} + 3 x ^{2} - 12 x - 20}{4} - 0

化简

= \frac{2 x ^{3} + 3 x ^{2} - 12 x - 20}{4}

x \to 3 + lim (\frac{x + 1}{x ^{2} - 9})

\int (8 x - 10 y + 2) d y

y^{^{''}} + 2 y^{^{'}} + 2 y = 0

\frac{\partial}{\partial y} (\frac{x}{x ^{6} - y ^{4}})

y^{^{'}} + y = e^{- t} - 1