tangent f(x)=x^2-x,(-4,20)

解答

切线

f (x) = x^{2} - x, (- 4, 20)

y = - 9 x - 16

求解步骤

计算

f (x) = x^{2} - x

的斜率:

\frac{df ( x )}{d x} = 2 x - 1

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 = - 9

找到斜率 m=

- 9

且穿过

(- 4, 20)

的直线:

y = - 9 x - 16

y = - 9 x - 16

x \cdot y \cdot \frac{d y}{d x} - y^{2} = 2 \cdot x^{2}

n = 4 \sum \infty \frac{1}{n ( n + 1 )}

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

\int_{0}^{2} \frac{1}{x + 3} d x

d er i v a t i v e cos (2 y)