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InterviewSolution
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Differentiate `{x^(tanx)+sqrt((x^(2)+1)/(x))}` w.r.t. x. |
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Answer» Let `y=u+v`, where `y=x^(tanx) and v=sqrt((x^(2)+1)/(x)).` Now, `u=x^(tanx)` `rArr logu=(tanx)(logx)` `rArr(1)/(u).(du)/(dx)=(tanx).(d)/(dx)(logx)+(logx).(d)/(dx)(tanx)` `" [differentiating w.r.t. x]"` `=(tanx).(1)/(x)+(logx)sec^(2)x` `rArr(du)/(dx)=u.[(tanx)/(x)+(logx)sec^(2)x]` `rArr(du)/(dx)=x^(tanx).{(tanx)/(x)+(logx)sec^(2)x}." ...(i)"` And, `v=sqrt((x^(2)+1)/(x))` `rArr log v=(1)/(2).{log (x^(2)+1)-logx}` `rArr(1)/(v).(dv)/(dx)=(1)/(2).{(2x)/((x^(2)+1))-(1)/(x)}" [differentiating w.r.t. x]"` `rArr(dv)/(dx)=(v)/(2).{(2x^(2)-(x^(2)+1))/(x(x^(2)+1))}` `rArr(dv)/(dx)=(1)/(2)sqrt((x^(2)+1)/(x)).{(x^(2)-1)/(x(x^(2)+1))}." ...(ii)"` `therefore y=u+v` `rArr(dy)/(dx)=(du)/(dx)+(dv)/(dx)` `rArr(dy)/(dx)=x^(tanx).{(tanx)/(x)+(logx)sec^(2)x}+(1)/(2).sqrt((x^(2)+1)/(x)).{((x^(2)-1))/(x(x^(2)+1))}.` |
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