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If x^x+y^x=1 , prove that (dy)/(dx)=-{(x...

If `x^x+y^x=1` , prove that `(dy)/(dx)=-{(x^x(1+logx)+y^x. logy)/(x . y^((x-1)))}`

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Given that,
`x^x+y^x=1`
Let `u=x^x` and `v=y^x`
`=>logu=xlogx`
`=>1/u(du)/(dx)=x xx 1/x +logx`
`=>(du)/(dx)=x^x(1+logx)`
Again `v=y^x`
`=>logv=xlogy`
`=>1/v(dv)/(dx)=x xx 1/y (dy)/(dx)+log y`
`=>(dv)/(dx)=y^x[xy^-1 (dy)/(dx) +logy]`
so,`x^x(1+logx)+y^x[xy^-1 (dy)/(dx) +logy]=0`
`=>x^x(1+logx)=-y^x[xy^-1 (dy)/(dx) +logy]`
`=>x^x(1+logx)=-xy^(x-1)(dy)/(dx)-y^xlogy`
`=>x^x(1+logx)+y^xlogy=-xy^(x-1)(dy)/(dx)`
`=>(dy)/(dx)=(x^x(1+logx)+y^xlogy)/(-xy^(x-1))`
`(dy)/(dx)=-{(x^x(1+logx)+y^x. logy)/(x . y^((x-1)))}`
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