tangent y=(5x)/(1+x^2),(5, 25/26)

解答

切线

y = \frac{5 x}{1 + x ^{2}}, (5, \frac{25}{26})

y = - \frac{30}{169} x + \frac{625}{338}

求解步骤

计算

y = \frac{5 x}{1 + x ^{2}}

的斜率:

\frac{d y}{d x} = \frac{5 ( 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}{169}

找到斜率 m=

- \frac{30}{169}

且穿过

(5, \frac{25}{26})

的直线:

y = - \frac{30}{169} x + \frac{625}{338}

y = - \frac{30}{169} x + \frac{625}{338}

\int w csc (4 w^{2} - 1) d w

\frac{d}{d x} (1.5 x - 0.25 x^{2})

l a pl a ce t r an s f or m cos (x)

\frac{d}{d x} (x e^{- x})

\int_{1}^{2} x^{\frac{1}{2}} d x