finding the 2nd derivative

Question

x+y=tany; show that y"= - (2cot^3y + 2cot^5y)

Answer 1

x+y=tan(y), 1+y'=y'sec²(y), y'(sec²(y)-1)=1,

y'tan²(y)=1, y'=cot²(y), y'²=cot⁴(y);

y"tan²(y)+2y'²tan(y)sec²(y)=0,

y"tan²(y)=-2y'²tan(y)sec²(y),

y"=-2y'²cot(y)(1+tan²(y))=-2y'²(cot(y)+tan(y))=-2(cot⁵(y)+cot³(y)),

which can be written y"=-(2cot³(y)+2cot⁵(y)) QED