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

解答

切线

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

y = \frac{13}{16} x - \frac{5}{8}

求解步骤

计算

(10 - x) y^{2} = x^{3}

的斜率:

\frac{d y}{d x} = \frac{3 x ^{2} + y ^{2}}{2 y ( 10 - 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{13}{16}

找到斜率 m=

\frac{13}{16}

且穿过

(2, 1)

的直线:

y = \frac{13}{16} x - \frac{5}{8}

y = \frac{13}{16} x - \frac{5}{8}

\int \frac{1}{16 x ^{2} + 4} d x

d er i v a t i v e x + (x - 6)^{6}

\frac{d}{d x} (e^{2 + x} - 2 e^{x} + 2 e x)

x \to \frac{π}{3} lim (cot (3 x))

\frac{\partial}{\partial x} (x^{2} e^{5 x y})