Question

如何用导数求曲线的切线方程?如圆.(x-a)^2+(y-b)^2=r^2,过点（x0,y0）(在圆上)的切线为(x0-a

Answer 1

(x-a)^2+(y-b)^2=r^2两边对x求导:

[(x-a)^2+(y-b)^2]' = (r^2)'

[(x-a)^2]' + [(y-b)^2]' = 0 (和的导数等于导数的和; 常数的导数为0）

2(x-a)*(x-a)' + 2(y-b)*(y-b)' = 0 (x^n的导数为nx^(n-1) )

2(x-a) + 2(y-b)y' = 0

y' = -(x-a)/(y-b)

点(x0,y0）处的切线斜率为:y' = -(x0 -a)/(y0 -b)

点斜式:y - y0 = [-(x0 -a)/(y0 -b)]*(x - x0)

(x0 -a) (x - x0) + (y0 - b)(y - y0) = 0

(x0 -a) (x - a + a - x0) + (y0 - b)(y - b + b- y0) = 0

展开:(x0 -a) (x - a) - (x0 - a)^2 + (y0 - b)(y - b) - (y0 - b)^2 = 0

(x0 -a) (x - a) - + (y0 - b)(y - b) = (x0 - a)^2 + (y0 - b)^2

点(x0,y0)在圆上,(x0 - a)^2 + (y0 - b)^2 = r^2

(x0-a)(x-a)+(y0-b)(y-b)=r^2

y'是y的导数(即dy/dx)