tangent f(x)=(5x)/(x+2),(3,3)

解答

切线

f (x) = \frac{5 x}{x + 2}, (3, 3)

y = \frac{2}{5} x + \frac{9}{5}

求解步骤

计算

f (x) = \frac{5 x}{x + 2}

的斜率:

\frac{df ( x )}{d x} = \frac{10}{( 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{2}{5}

找到斜率 m=

\frac{2}{5}

且穿过

(3, 3)

的直线:

y = \frac{2}{5} x + \frac{9}{5}

y = \frac{2}{5} x + \frac{9}{5}

\int_{0}^{4} \frac{x}{8} d x

t an g e n t f (x) = e^{x} - x e^{x}, at x = 0

x \to - 11 lim (x^{2} + 11)

x \to 0 lim (arctan (e^{x}))

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