tangent 9y^7+4x^3=8y+5x,(1,1)

解答

切线

9 y^{7} + 4 x^{3} = 8 y + 5 x, (1, 1)

y = - \frac{7}{55} x + \frac{62}{55}

求解步骤

计算

9 y^{7} + 4 x^{3} = 8 y + 5 x

的斜率:

\frac{d y}{d x} = \frac{5 - 12 x ^{2}}{63 y ^{6} - 8}

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{7}{55}

找到斜率 m=

- \frac{7}{55}

且穿过

(1, 1)

的直线:

y = - \frac{7}{55} x + \frac{62}{55}

y = - \frac{7}{55} x + \frac{62}{55}

\frac{\partial}{\partial x} (cos^{5} (x^{7} y^{5}))

x y^{^{'}} + (1 - x) y = 0

\frac{d N}{d t} = N (1 - 0.0005 N)

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

\frac{d}{d x} (y - x n (y + x))