求导和求法线问题

Question

1运用极限的定义求的导数。
2求在点（2，√6）上的法线

Answer 1

f'(x)=lim h->0 [f(x+h)-f(x)]/h
=lim h->0 {(x+h-1)/[(x+h)^2+3(x+h)]-(x-1)/(x^2+3x)}/h
(x+h-1)(x^2+3x)-(x-1)[(x+h)^2+3(x+h)]
=lim h->0 ------------------------------------------------
h[(x+h)^2+3(x+h)](x^2+3x)
h[-x^2+(2-h)x+h+3]
=lim h->0 ------------------------------------------------
h[(x+h)^2+3(x+h)](x^2+3x)
消去h，取极限
-x^2+2x+3
= ------------------------
(x^2+3x)^2

记住法线和切线垂直，过同一定点
先求切线斜率
k切=f'(x)
=[(x^3-x)^(1/2)]'
=(1/2)(x^3-x)^(-1/2)*(x^3-x)'
=(1/2)(x^3-x)^(-1/2)*(3x^2-1)
f'(2)=(1/2)(6)^(-1/2)*(3*4-1)
=11/(2根号6)
k法=-1/k切=-2根号6/11
且过点(2,根号6)
所以法线方程为
y-根号6=(-2根号6/11)(x-2)