tangent y=sqrt(-3x-9),(-9,sqrt(18))

解答

切线

y = - 3 x - 9, (- 9, 18)

y = - \frac{1}{2 2} x + 32 - \frac{9}{2 2}

求解步骤

计算

y = - 3 x - 9

的斜率:

\frac{d y}{d x} = - \frac{3}{2 - 3 x - 9}

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 2}

找到斜率 m=

- \frac{1}{2 2}

且穿过

(- 9, 18)

的直线:

y = - \frac{1}{2 2} x + 32 - \frac{9}{2 2}

y = - \frac{1}{2 2} x + 32 - \frac{9}{2 2}

d er i v a t i v e g (x) = 2 x^{0.8} + 4 x^{- 0.8}

\int 3 x^{2} + 2 y^{2} d x

d er i v a t i v e f (x) = cot^{2} (cos (x))

t an g e n t y = \frac{2}{x - 6}

d er i v a t i v e y = - \frac{2 x ^{2}}{5 y ^{2}}