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受欢迎的微积分 >

tangent 2x(x+1)^3(x^2+2x+1)^2

解答

切线

2 x (x + 1)^{3} (x^{2} + 2 x + 1)^{2}

解答

y = 2 (8 a_{0}^{7} + 49 a_{0}^{6} + 126 a_{0}^{5} + 175 a_{0}^{4} + 140 a_{0}^{3} + 63 a_{0}^{2} + 14 a_{0} + 1) x - 14 a_{0}^{8} - 84 a_{0}^{7} - 210 a_{0}^{6} - 280 a_{0}^{5} - 210 a_{0}^{4} - 84 a_{0}^{3} - 14 a_{0}^{2}

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

(a_{0}, 2 a_{0} (a_{0} + 1)^{3} (a_{0}^{2} + 2 a_{0} + 1)^{2})

计算

y = 2 x (x + 1)^{3} (x^{2} + 2 x + 1)^{2}

的斜率:

\frac{d y}{d x} = 2 ((x + 1)^{3} (x^{2} + 2 x + 1)^{2} + x (3 (x + 1)^{2} (x^{2} + 2 x + 1)^{2} + 2 (x^{2} + 2 x + 1) (2 x + 2) (x + 1)^{3}))

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2 (8 a_{0}^{7} + 49 a_{0}^{6} + 126 a_{0}^{5} + 175 a_{0}^{4} + 140 a_{0}^{3} + 63 a_{0}^{2} + 14 a_{0} + 1)

找到斜率 m=

2 (8 a_{0}^{7} + 49 a_{0}^{6} + 126 a_{0}^{5} + 175 a_{0}^{4} + 140 a_{0}^{3} + 63 a_{0}^{2} + 14 a_{0} + 1)

且穿过

(a_{0}, 2 a_{0} (a_{0} + 1)^{3} (a_{0}^{2} + 2 a_{0} + 1)^{2})

的直线:

y = 2 (8 a_{0}^{7} + 49 a_{0}^{6} + 126 a_{0}^{5} + 175 a_{0}^{4} + 140 a_{0}^{3} + 63 a_{0}^{2} + 14 a_{0} + 1) x - 14 a_{0}^{8} - 84 a_{0}^{7} - 210 a_{0}^{6} - 280 a_{0}^{5} - 210 a_{0}^{4} - 84 a_{0}^{3} - 14 a_{0}^{2}

y = 2 (8 a_{0}^{7} + 49 a_{0}^{6} + 126 a_{0}^{5} + 175 a_{0}^{4} + 140 a_{0}^{3} + 63 a_{0}^{2} + 14 a_{0} + 1) x - 14 a_{0}^{8} - 84 a_{0}^{7} - 210 a_{0}^{6} - 280 a_{0}^{5} - 210 a_{0}^{4} - 84 a_{0}^{3} - 14 a_{0}^{2}

作图

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