tangent f(x)= 3/(sqrt(x)),(1,3)

解答

切线

f (x) = \frac{3}{x}, (1, 3)

y = - \frac{3}{2} x + \frac{9}{2}

求解步骤

计算

f (x) = \frac{3}{x}

的斜率:

\frac{df ( x )}{d x} = - \frac{3}{2 x ^{\frac{3}{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{3}{2}

找到斜率 m=

- \frac{3}{2}

且穿过

(1, 3)

的直线:

y = - \frac{3}{2} x + \frac{9}{2}

y = - \frac{3}{2} x + \frac{9}{2}

y^{^{''}} + 3 y^{^{'}} - 10 y = x (e^{x} + 1)

\frac{d}{d x} (3^{- 6 x + 6})

f^{^{'}} (x) = 3 x^{2} (8 - x^{- 3})

\int_{0}^{1} 5^{2 x} d x

\int_{0}^{1} x^{3} + x d x