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tangent y=arctan(-x),\at x=3

解答

切线

y = arctan (- x), at x = 3

解答

y = - \frac{1}{10} x - arctan (3) + \frac{3}{10}

求解步骤

找到切点:

(3, - arctan (3))

计算

y = arctan (- x)

的斜率:

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

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}{10}

找到斜率 m=

- \frac{1}{10}

且穿过

(3, - arctan (3))

的直线:

y = - \frac{1}{10} x - arctan (3) + \frac{3}{10}

y = - \frac{1}{10} x - arctan (3) + \frac{3}{10}

作图

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