tangent y=(2x+1)/(x+2),(1,1)

解答

切线

y = \frac{2 x + 1}{x + 2}, (1, 1)

y = \frac{1}{3} x + \frac{2}{3}

求解步骤

计算

y = \frac{2 x + 1}{x + 2}

的斜率:

\frac{d y}{d x} = \frac{3}{( x + 2 ) ^{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{1}{3}

找到斜率 m=

\frac{1}{3}

且穿过

(1, 1)

的直线:

y = \frac{1}{3} x + \frac{2}{3}

y = \frac{1}{3} x + \frac{2}{3}

\frac{\partial}{\partial y} (cos (2 x y))

d er i v a t i v e g (x) = 9 - 2

\int_{- 1}^{1} \frac{5 x}{( x + 3 ) ^{2}} d x

\int \frac{1}{1 + 4 y ^{2}} d y

x \to - 4 lim (7^{x})