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tangent y= x/((1+x^2),\at (3,0.3))

解答

切线

y = \frac{x}{( 1 + x ^{2} ) , at ( 3 , 0.3 )}

解答

y = \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{a _{0}}{( 1 + a _{0}^{2} ) , ( 3 , 0.3 )} - \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}} a_{0}

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}}{( 1 + a _{0}^{2} ) , ( 3 , 0.3 )})

计算

y = \frac{x}{( 1 + x ^{2} ) , ( 3 , 0.3 )}

的斜率:

\frac{d y}{d x} = \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + x ^{2} ) - 2 xx}{( 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{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}}

找到斜率 m=

\frac{1}{, ( 3 , 0.3 )} \frac{1 ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}}

且穿过

(a_{0}, \frac{a _{0}}{( 1 + a _{0}^{2} ) , ( 3 , 0.3 )})

的直线:

y = \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{a _{0}}{( 1 + a _{0}^{2} ) , ( 3 , 0.3 )} - \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}} a_{0}

y = \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{a _{0}}{( 1 + a _{0}^{2} ) , ( 3 , 0.3 )} - \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}} a_{0}

作图

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