tangent 2xy^3+xy=18,(6,1)

解答

切线

2 x y^{3} + x y = 18, (6, 1)

y = - \frac{1}{14} x + \frac{10}{7}

求解步骤

计算

2 x y^{3} + x y = 18

的斜率:

\frac{d y}{d x} = \frac{- y - 2 y ^{3}}{x ( 1 + 6 y ^{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}{14}

找到斜率 m=

- \frac{1}{14}

且穿过

(6, 1)

的直线:

y = - \frac{1}{14} x + \frac{10}{7}

y = - \frac{1}{14} x + \frac{10}{7}

l a pl a ce t r an s f or m \frac{1}{e ^{j t}}

\frac{d}{d x} (e^{x} (x - 9))

\frac{d}{d x} (\frac{x ^{5} + 5}{x ^{3}})

a re a 4 x^{2} + 10, 2 x + 7, - 3, 2

\frac{d}{d x} (3 x^{2} + \frac{1}{x ^{2}})