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tangent y=x^{1/3}

解答

切线

y = x^{\frac{1}{3}}

解答

y = \frac{1}{3 a _{0}^{\frac{2}{3}}} x + a_{0}^{\frac{1}{3}} - \frac{a _{0}^{\frac{1}{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}^{\frac{1}{3}})

计算

y = x^{\frac{1}{3}}

的斜率:

\frac{d y}{d x} = \frac{1}{3 x ^{\frac{2}{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{1}{3 a _{0}^{\frac{2}{3}}}

找到斜率 m=

\frac{1}{3 a _{0}^{\frac{2}{3}}}

且穿过

(a_{0}, a_{0}^{\frac{1}{3}})

的直线:

y = \frac{1}{3 a _{0}^{\frac{2}{3}}} x + a_{0}^{\frac{1}{3}} - \frac{a _{0}^{\frac{1}{3}}}{3}

y = \frac{1}{3 a _{0}^{\frac{2}{3}}} x + a_{0}^{\frac{1}{3}} - \frac{a _{0}^{\frac{1}{3}}}{3}

作图

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