tangent 2x^2+3xy+1y^3=-10,(-3,-4)

解答

切线

2 x^{2} + 3 x y + 1 y^{3} = - 10, (- 3, - 4)

y = \frac{8}{13} x - \frac{28}{13}

求解步骤

计算

2 x^{2} + 3 x y + 1 y^{3} = - 10

的斜率:

\frac{d y}{d x} = \frac{- 4 x - 3 y}{3 ( y ^{2} + x )}

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{8}{13}

找到斜率 m=

\frac{8}{13}

且穿过

(- 3, - 4)

的直线:

y = \frac{8}{13} x - \frac{28}{13}

y = \frac{8}{13} x - \frac{28}{13}

x^{2} y^{^{''}} + x y^{^{'}} + y = sec (ln (x))

\frac{d}{d x} (\frac{x + 2}{x - 3})

\frac{d}{d x} (ln^{3} (3 x^{2} - 5))

\int (x^{2} + 2)^{3} d x

x \to 0 lim ((x - 4)^{3})