derivative (t+3)(2t-1)^2

解答

导数

(t + 3) (2 t - 1)^{2}

12 t^{2} + 16 t - 11

求解步骤

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

使用微分乘法定则:

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

f = t + 3, g = (2 t - 1)^{2}

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

\frac{d}{d t} (t + 3) = 1

\frac{d}{d t} ((2 t - 1)^{2}) = 4 (2 t - 1)

= 1 \cdot (2 t - 1)^{2} + 4 (2 t - 1) (t + 3)

化简

1 \cdot (2 t - 1)^{2} + 4 (2 t - 1) (t + 3) : 12 t^{2} + 16 t - 11

= 12 t^{2} + 16 t - 11

n = 0 \sum \infty n \cdot (\frac{4}{5})^{n}

\frac{d ^{2}}{d x ^{2}} ((3 + 2 x) e^{- 5 x})

t an g e n t f (x) = x^{2} + 1, at x = 1

x \to 2 lim (3 x^{2} - 5 x + 1)

\int_{0}^{1} x^{5} d x