斜率 (6,-4),(3,1)

解答

斜率

(6, - 4), (3, 1)

m = - \frac{5}{3}

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}) = (6, - 4), (x_{2}, y_{2}) = (3, 1)

m = \frac{1 - ( - 4 )}{3 - 6}

整理后得

m = - \frac{5}{3}

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

t an g e n t f (x) = 6 tan (x)

x \to 2 lim (\frac{1}{( x - 2 ) ^{2}})

d er i v a t i v e w (z) = 6 z^{2} e^{z}

\int_{0}^{l n (2)} \int_{1}^{l n (5)} e^{2 x + y} d y d x