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tangent y=x^3-3x+1

解答

切线

y = x^{3} - 3 x + 1

解答

y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} + 1

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} - 3 a_{0} + 1)

计算

y = x^{3} - 3 x + 1

的斜率:

\frac{d y}{d x} = 3 x^{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 = 3 a_{0}^{2} - 3

找到斜率 m=

3 a_{0}^{2} - 3

且穿过

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

的直线:

y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} + 1

y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} + 1

作图

流行的例子

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\frac{d}{d x} (lo g_{3} (sin (x^{2})))

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x \to - 2 + lim (2 - x)

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\int x^{2} x + 5 d x

integral of iny

\int in y d y

(1+x^4)(dy}{dx}=\frac{x^3)/y

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