Differentiate the following w.r.t. x:
`(i) sec(tan^(-1)x)" "(ii)sin(tan^(-1)x)" "(iii)cot(cos^(-1)x)`

Explanation

(i) Let `y=sec(tan^(-1)x).`
Putting `tan^(-1)x=t,` we get `y=sec t and t = tan^(-1)x`.
Now `y=sec t rArr (dy)/(dt)=sec t tan t.`
And, `t= tan^(-1) x rArr (dt)/(dx)=(1)/((1+x^(2))).`
`therefore(dy)/(dx)=((dy)/(dt)xx(dt)/(dx))=(sec t tan t)/((1+x^(2)))=((sqrt(1+tan^(2)t))(tan t))/((1+x^(2)))`
`=((sqrt(1+x^(2)))x)/((1+x^(2)))=(x)/(sqrt(1+x^(2)))" "[because t = tan^(-1)x rArr tan t = x]`
Hence, `(d)/(dx){sec(tan^(-1)x)}=(x)/(sqrt(1+x^(2))).`
(ii) Let `y=sin (tan^(-1)x)`
Putting `tan^(-1)x=t`, we get `y=sin t and t= tan^(-1)x.`
Now, `y=sin t rArr (dy)/(dt)=cos t.`
And, `t=tan^(-1)x rArr (dt)/(dx)=(1)/((1+x^(2)))`.
`therefore(dy)/(dx)=((dy)/(dt)xx(dt)/(dx))=cost.(1)/((1+x^(2)))=(1)/((1+x^(2))^(3//2))`
`" "[because tan t = x rArr cot t = (1)/(sqrt(1+x^(2)))].`
Hence, `(d)/(dx)[sin(tan^(-1)x)}=(1)/((1+x^(2))^(3//2)).`
(iii) Let `y=cot(cos^(-1)x)`
Putting `cos^(-1)x=t`, we get `y=cot t and t = cos^(-1)x.`
Now, `y=cot t rArr (dy)/(dt)=-"cosec"^(2)t.`
And, `t=cos^(-1)xrArr (dt)/(dx)=(-1)/(sqrt(1-x^(2))).`
`therefore(dy)/(dx)=((dy)/(dt)xx(dt)/(dx))=("cosec"^(2)t)/(sqrt(1+x^(2)))=(1)/((1+x^(2))^(3//2))`
`" "[because cos t = x rArr "cosec t" = (1)/(sqrt(1+x^(2)))].`
Hence, `(d)/(dx){cot^(-1)(cos^(-1) x)}=(1)/((1-x^(2))^(3//2)).`