InterviewSolution
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InterviewSolution
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If `x^y+y^x=a^b` , then find `dy/dx`. |
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Answer» Let `u=x^(y) and v=y^(x)`. Then, `u+v=a^(b)rArr(du)/(dx)+(dv)/(dx)=0." ...(i) "[becausea^(b)=" constant"]` Now, `u=x^(y)rArrlogu=ylogx" [taking log on both sides]"` `rArr(1)/(u).(du)/(dx)=y.(1)/(x)+logx.(dy)/(dx)" [on differentiaton]"` `rArr(du)/(dx)=u[(y)/(x)+logx.(dy)/(dx)]` `rArr(du)/(dx)=x^(y)[(y+xlogx.(dy)/(dx))/(x)]` `rArr(du)/(dx)=x^(y-1)[y+xlogx.(dy)/(dx)].` And, `v=y^(x)rArr log v=x logy" [taking log on both sides]"` `rArr(1)/(v).(dv)/(dx)=x.(1)/(y).(dy)/(dx)+logy" [on differentiation]"` `rArr(dv)/(dx)=v.[(x)/(y).(dy)/(dx)+logy]rArr(dv)/(dx)=y^(x){(x.(dy)/(dx)+ylogy)/(y)}` `rArr(dv)/(dx)=y^((x-1)).{x.(dy)/(dx)+ylogy}.` Using (i), we get `(du)/(dx)+(dv)/(dx)=0` `rArrx^((y-1)){y+xlogx.(dy)/(dx)}+y^((x-1)).{x(dy)/(dx)+ylogy}=0` `rArr{x^(y)(logx)+x.y^((x-1))}.(dy)/(dx)=-{y.x^((y-1))+y^(x)(logy)}.` `therefore(dy)/(dx)=(-{y.x^((y-1))+y^(x)(logy)})/({x^(y)(logx)+xy^((x-1))}).` |
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