tangent xy^8+8xy=36,(4,1)

解答

切线

x y^{8} + 8 x y = 36, (4, 1)

y = - \frac{9}{64} x + \frac{25}{16}

求解步骤

计算

x y^{8} + 8 x y = 36

的斜率:

\frac{d y}{d x} = \frac{- y ^{8} - 8 y}{8 x ( y ^{7} + 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{9}{64}

找到斜率 m=

- \frac{9}{64}

且穿过

(4, 1)

的直线:

y = - \frac{9}{64} x + \frac{25}{16}

y = - \frac{9}{64} x + \frac{25}{16}

\frac{d}{d x} (\frac{13}{3} x^{3} + \frac{21}{2} x^{2} - 10 x - 7)

\int \frac{x ^{2}}{3 + 1} d x

t an g e n t f (x) = sin (sin (x)), (3 π, 0)

d er i v a t i v e f (x) = x - 3

\int 2 y x d x