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

解答

切线

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

y = - \frac{1}{2} x + \frac{3}{2}

求解步骤

计算

x y^{3} + x y = 2

的斜率:

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

找到斜率 m=

- \frac{1}{2}

且穿过

(1, 1)

的直线:

y = - \frac{1}{2} x + \frac{3}{2}

y = - \frac{1}{2} x + \frac{3}{2}

t an g e n t f (x) = \frac{4}{x ^{2}}, at x = 1

\frac{d}{d x} (\frac{5 x ^{3} + 7 x ^{2}}{x})

\frac{d}{d x} (x - \frac{1}{9} x)

x \to 0 lim (e)

n = 1 \sum \infty \frac{n - 1}{3 n - 1}