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受欢迎的微积分 >

tangent f(x)= 4/(1+x^2)

解答

切线

f (x) = \frac{4}{1 + x ^{2}}

解答

y = - \frac{8 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{12 a _{0}^{2} + 4}{( a _{0}^{2} + 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{4}{1 + a _{0}^{2}})

计算

f (x) = \frac{4}{1 + x ^{2}}

的斜率:

\frac{df ( x )}{d x} = - \frac{8 x}{( 1 + x ^{2} ) ^{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{8 a _{0}}{( 1 + a _{0}^{2} ) ^{2}}

找到斜率 m=

- \frac{8 a _{0}}{( 1 + a _{0}^{2} ) ^{2}}

且穿过

(a_{0}, \frac{4}{1 + a _{0}^{2}})

的直线:

y = - \frac{8 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{12 a _{0}^{2} + 4}{( a _{0}^{2} + 1 ) ^{2}}

y = - \frac{8 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{12 a _{0}^{2} + 4}{( a _{0}^{2} + 1 ) ^{2}}

作图

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