InterviewSolution
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InterviewSolution
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If `y=(logx)^(cosx)+(x^(2)+1)/(x^(2)-1)`, find `(dy)/(dx).` |
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Answer» Let `(logx)^(cosx)=u and(x^(2)+1)/(x^(2)-1)=v.` Then, `u=(logx)^(cosx)rArr log u = cosx. Log (logx)." …(i)"` On differentiating both sides of (i) w.r.t. x, we get `(1)/(u).(du)/(dx)=cosx.(d)/(dx){log(logx)}+log(logx).(d)/(dx)(logx)` `=cosx(1)/(logx).(1)/(x)-(sinx).log(logx).` `therefore(du)/(dx)=u.{(cosx)/(xlogx)-(sinx).log(logx)}` `rArr(du)/(dx)=(logx)^(cosx).{(cosx)/(x logx)-(sinx).log(logx)}.` And, `v=((x^(2)+1))/((x^(2)-1))rArr(dv)/(dx)=((x^(2)-1).(d)/(dx)(x^(2)+1)-(x^(2)+1).(d)/(dx)(x^(2)-1))/((x^(2)-1)^(2))` `rArr(dv)/(dx)=((x^(2)-1).2x-(x^(2)+1).2x)/((x^(2)-1)^(2))=(-4x)/((x^(2)-1)^(2)).` `therefore y=u+v` `rArr(dy)/(dx)=(du)/(dx)+(dv)/(dx)` `=(logx)^(cosx).{(cosx)/(xlogx)-(sinx).(logx)}-(4x)/((x^(2)-1)^(2)).` |
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