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tangent f(x)= x/(1+x^2),0.4,\at x=2

解答

切线

f (x) = \frac{x}{1 + x ^{2}}, 0.4, at x = 2

解答

y = - \frac{3}{25} x + \frac{16}{25}

求解步骤

找到切点:

(2, \frac{2}{5})

计算

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

的斜率:

\frac{df ( x )}{d x} = \frac{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{3}{25}

找到斜率 m=

- \frac{3}{25}

且穿过

(2, \frac{2}{5})

的直线:

y = - \frac{3}{25} x + \frac{16}{25}

y = - \frac{3}{25} x + \frac{16}{25}

作图

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