tangent-3x^2+2xy+2y^3=125,(3,4)

解答

切线

- 3 x^{2} + 2 x y + 2 y^{3} = 125, (3, 4)

y = \frac{5}{51} x + \frac{63}{17}

求解步骤

计算

- 3 x^{2} + 2 x y + 2 y^{3} = 125

的斜率:

\frac{d y}{d x} = \frac{3 x - y}{x + 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{5}{51}

找到斜率 m=

\frac{5}{51}

且穿过

(3, 4)

的直线:

y = \frac{5}{51} x + \frac{63}{17}

y = \frac{5}{51} x + \frac{63}{17}

\frac{d}{d x} y = ln (10 x^{2} - 8)

\frac{d}{d x} (I n x)

t an g e n t f (x) = x^{3} - 2 x^{2} + 1, at x = 1

\frac{d}{d x} (arcsin (e^{2 x}))

\int \frac{x ^{4}}{5} d x