Question

求一个隐函数的二阶导数

y=tan（x+y）求个详细步骤，谢谢

Answer 1

利用隐函数的微分法求解：令F（x,y(x))=0.两边对x求导，得：dF/dx+(dF/dy)*(dy/dx)=0.若dF/dy<>0,则dy/dx=-(dF/dx)/(dF/dy).于是题目可以这样解：设F=y-tan(x+y),dF/dx=-sec²（x+y),dF/dy=1-sec²（x+y)=-tan²（x+y),所以dy/dx=-sec²（x+y)/tan²（x+y)=-csc²(x+y).

Answer 2

解： y'=(sec(x+y))^2 * (1+y')
y'=(sec(x+y))^2 + y'*(sec(x+y))^2
y'*(1-(sec(x+y))^2)=(sec(x+y))^2
y'=(sec(x+y))^2 / ( 1-(sec(x+y))^2)