InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
If `(tan^(-1)x)^(y)+y^(cotx)=1,` then find `(dy)/(dx).` |
|
Answer» Let `(tan^(-1)x)^(y)=u and y^(cotx)=v`. Then, `u+v=1 rArr (du)/(dx)+(dv)/(dx)=0." …(i)"` Now, `u=(tan^(-1)x)^(y)` `rArr logu=ylog (tan^(-1)x)` `rArr(1)/(u).(du)/(dx)=y.(1)/(tan^(-1)x).(1)/((1+x^(2)))+log(tan^(-1)x)(dy)/(dx)` `rArr (du)/(dx)=u.{(y)/((1+x^(2)tan^(-1)x))+log(tan^(-1)x).(dy)/(dx)}` `rArr(du)/(dx)=(tan^(-1)x)^(y).{(y)/((1+x^(2))tan^(-1)x)+log(tan^(-1)x).(dy)/(dx)}` `rArr (du)/(dx)=(y(tan^(-1)x)^(y-1))/((1+x^(2)))+(tan^(-1)x)^(y)log(tan^(-1)x).(dy)/(dx)." ...(ii)"` Again, `v=y^(cotx)` `rArr logv=(cotx)logy` `rArr (1)/(v).(dv)/(dx)=(cotx).(1)/(y).(dy)/(dx)+(logy)(-"cosec"^(2)x)` `rArr(dv)/(dx)=v{(cotx)/(y).(dy)/(dx)-("cosec"^(2)x)(logy)}` `rArr (dv)/(dx)=y^(cotx){(cotx)/(y).(dy)/(dx)-("cosec"^(2)x)(logy)}` `rArr (dv)/(dx)=(cotx)y^((cotx)-1).(dy)/(dx)-y^(cotx)("cosec"^(2)x)(logy)." ...(iii)"` Using (ii) and (iii) in (i), we get `(y(tan^(-1)x)^(y-1))/((1+x^(2)))+(tan^(-1)x)^(y)log(tan^(-1)x)(dy)/(dx)+(cotx)y^((cotx)-1).(dy)/(dx)-y^(cotx)("cosec"^(2)x)(logy)=0` `rArr{(tan^(-1)x)^(y)log(tan^(-1)x)+(cotx)y^((cotx)-1)}(dy)/(dx)={y^(cotx)("cosec"^(2)x)(logy)-(y(tan^(-1)x)^(y-1))/((1+x^(2)))}` `rArr(dy)/(dx)=(y^(cotx)("cosec"^(2)x)(logy)-(y(tan^(-1)x)^(y-1))/((1+x^(2))))/((tan^(-1)x)^(y)log(tan^(-1)x)+(cotx)y^((cotx)-1)).` |
|