Differentiate y=cot^-1(t)+cot^-1(1/t). // y'=.....?

Question

Differentiate y=cot^-1(t)+cot^-1(1/t).

y'=.....?

Answer 1

For x>0 we have cot^-1(1/x)=tan^-1(x). Therefore:

y=cot^-1(x)+cot^-1(1/x)=cot^-1(x)+tan^-1(x)=π/2

y'=0

For x<0 we have cot^-1(1/x)=tan^-1(x)+π. Therefore:

y=cot^-1(x)+cot^-1(1/x)=cot^-1(x)+tan^-1(x)+π=3π/2

y'=0

Therefore, for all x we have:

y'=0