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tangent x^3sqrt(x^3+3)

解答

切线

x^{3} x^{3} + 3

解答

y = \frac{9 a _{0}^{5} + 18 a _{0}^{2}}{2 a _{0}^{3} + 3} x + a_{0}^{3} a_{0}^{3} + 3 - \frac{a _{0} ( 9 a _{0}^{5} + 18 a _{0}^{2} )}{2 a _{0}^{3} + 3}

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} a_{0}^{3} + 3)

计算

y = x^{3} x^{3} + 3

的斜率:

\frac{d y}{d x} = \frac{9 x ^{5} + 18 x ^{2}}{2 x ^{3} + 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 = \frac{9 a _{0}^{5} + 18 a _{0}^{2}}{2 a _{0}^{3} + 3}

找到斜率 m=

\frac{9 a _{0}^{5} + 18 a _{0}^{2}}{2 a _{0}^{3} + 3}

且穿过

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

的直线:

y = \frac{9 a _{0}^{5} + 18 a _{0}^{2}}{2 a _{0}^{3} + 3} x + a_{0}^{3} a_{0}^{3} + 3 - \frac{a _{0} ( 9 a _{0}^{5} + 18 a _{0}^{2} )}{2 a _{0}^{3} + 3}

y = \frac{9 a _{0}^{5} + 18 a _{0}^{2}}{2 a _{0}^{3} + 3} x + a_{0}^{3} a_{0}^{3} + 3 - \frac{a _{0} ( 9 a _{0}^{5} + 18 a _{0}^{2} )}{2 a _{0}^{3} + 3}

作图

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