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

解答

切线

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

解答

y = - \frac{30}{289} x + \frac{256}{289}

求解步骤

找到切点:

(4, \frac{8}{17})

计算

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

的斜率:

\frac{df ( x )}{d x} = \frac{2 ( 1 - x ^{2} )}{( 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{30}{289}

找到斜率 m=

- \frac{30}{289}

且穿过

(4, \frac{8}{17})

的直线:

y = - \frac{30}{289} x + \frac{256}{289}

y = - \frac{30}{289} x + \frac{256}{289}

作图

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