Question

求由这个方程y=tan(x+y)

所确定的隐函数的二阶导数d^2y/dx^2 答案是-2csc^2(x+y)*cot^3(x+y)麻烦要过程

Answer 1

y=tan(x+y)
y'=[sec(x+y)]^2*(1+y')
则
y'=[sec(x+y)]^2/{1-[sec(x+y)]^2}=-[sec(x+y)]^2/tan(x+y)]^2=-1/[sin(x+y)]^2
则y''={-1/[sin(x+y)]^2}'={-2[sin(x+y)]cos(x+y)}*（1+y')1/[sin(x+y)]^4
则
y''={-2[sin(x+y)]cos(x+y)}（1-1/[sin(x+y)]^2)*（1+y')1/[sin(x+y)]^4
={-2[sin(x+y)]cos(x+y)}[cos(x+y)]^2*1/[sin(x+y)]^4
化简下去就行了
真麻烦！