斜率 (11/6 ,-1/2),(1/3 , 1/3)

解答

斜率

(\frac{11}{6}, - \frac{1}{2}), (\frac{1}{3}, \frac{1}{3})

m = - \frac{5}{9}

求解步骤

两点之间的斜率: 斜率

= \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

(x_{1}, y_{1}) = (\frac{11}{6}, - \frac{1}{2}), (x_{2}, y_{2}) = (\frac{1}{3}, \frac{1}{3})

m = \frac{\frac{1}{3} - ( - \frac{1}{2} )}{\frac{1}{3} - \frac{11}{6}}

整理后得

m = - \frac{5}{9}

s l o p e (1) (3.5)

\int 19 arctan (x) d x

f (x) = (4 x^{2} - 6 x)^{3}

d er i v a t i v e y = \frac{7 x ^{3} - 3 x ^{2} - 9}{x ^{8}}

\frac{d}{d x} (cos^{2} (x) \cdot sin (x))