Differentiate w.r.t. x :
`(i)cot^(-1)((1)/(x))" "(ii)tan^(-1)((2x)/(1-x^(2)))" "(iii)cot^(-1)((1-x)/(1+x))`

Explanation

(i) Let `y=cot^(-1)((1)/(x)).`
Putting `x= tan theta`, we get
`y=cot^(-1)((1)/(tan theta))=cot^(-1)(cot theta)=theta=tan^(-1)x.`
`therefore(dy)/(dx)=(1)/((1+x^(2))).`
Hence, `(d)/(dx){cot^(-1)((1)/(2))}=(1)/((1+x^(2))).`
(ii) Let `y=tan^(-1)((2x)/(1-x^(2))).`
Putting `x=tan theta`, we get
`y=tan^(-1)((2tan theta)/(1-tan^(2)theta))=tan^(-1)(tan2 theta)=2theta=2 tan^(-1)x.`
`therefore(dy)/(dx)=(2)/((1+x^(2))).`
Hence, `(d)/(dx){tan^(-1)((2x)/(1-x^(2)))}=(2)/((1+x^(2))).`
(iii) Let `y=cot^(-1)((1-x)/(1+x)).`
Putting `x=tan theta,` we get
`y=cot^(-1)((1-tan theta)/(1+tan theta))=cot^(-1){tan((pi)/(4)-theta)}`
`=cot^(-1)[cot{(pi)/(2)-((pi)/(4)-theta)}]=((pi)/(4)+theta)=(pi)/(4)+tan^(-1)x.`
`therefore (dy)/(dx)=(1)/((1+x^(2)))`
Hence, `(d)/(dx){cot^(-1)((1-x)/(1+x))}=(1)/((1+x^(2))).`