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tangent (x^3+2)^5

解答

切线

(x^{3} + 2)^{5}

解答

y = 15 a_{0}^{2} (a_{0}^{3} + 2)^{4} x - 14 a_{0}^{15} - 110 a_{0}^{12} - 320 a_{0}^{9} - 400 a_{0}^{6} - 160 a_{0}^{3} + 32

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

(a_{0}, (a_{0}^{3} + 2)^{5})

计算

y = (x^{3} + 2)^{5}

的斜率:

\frac{d y}{d x} = 15 x^{2} (x^{3} + 2)^{4}

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 = 15 a_{0}^{2} (a_{0}^{3} + 2)^{4}

找到斜率 m=

15 a_{0}^{2} (a_{0}^{3} + 2)^{4}

且穿过

(a_{0}, (a_{0}^{3} + 2)^{5})

的直线:

y = 15 a_{0}^{2} (a_{0}^{3} + 2)^{4} x - 14 a_{0}^{15} - 110 a_{0}^{12} - 320 a_{0}^{9} - 400 a_{0}^{6} - 160 a_{0}^{3} + 32

y = 15 a_{0}^{2} (a_{0}^{3} + 2)^{4} x - 14 a_{0}^{15} - 110 a_{0}^{12} - 320 a_{0}^{9} - 400 a_{0}^{6} - 160 a_{0}^{3} + 32

作图

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