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tangent y=(sqrt(x))/(x+1)

解答

切线

y = \frac{x}{x + 1}

解答

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

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

(a_{0}, \frac{a _{0}}{a _{0} + 1})

计算

y = \frac{x}{x + 1}

的斜率:

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

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{- a _{0} + 1}{2 a _{0} ( a _{0} + 1 ) ^{2}}

找到斜率 m=

\frac{- a _{0} + 1}{2 a _{0} ( a _{0} + 1 ) ^{2}}

且穿过

(a_{0}, \frac{a _{0}}{a _{0} + 1})

的直线:

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

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

作图

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