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If (sinx)^y=x+y , prove that (dy)/(dx)=(...

If `(sinx)^y=x+y` , prove that `(dy)/(dx)=(1-(x+y)ycotx)/((x+y)logsinx-1)`

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Given that,
`(sinx)^y=x+y`
Let `u=(sinx)^y`
`=>log u=y log(sinx)`
`=>d/(dx)log u=d/(dx)y log(sinx)`
`=>1/u (du)/(dx)=(dy)/(dx) * log(sinx) +y*d/(dx)[log(sinx)]`
`=>(du)/(dx)=u[(dy)/(dx) * log(sinx) +y*cotx]`
`=>(du)/(dx)=(sinx)^y[(dy)/(dx) * log(sinx) +y*cotx]`
Let `u=x+y`
`=>(du)/(dx)=1+(dy)/(dx)`
`=>(sinx)^y*(dy)/(dx) * log(sinx) +(sinx)^y*y*cotx=1+(dy)/(dx)`
`=>[(sinx)^y* log(sinx)-1](dy)/(dx)=1-(sinx)^y*y*cotx`
`=>(dy)/(dx)=(1-(sinx)^y *ycotx)/((sinx)^y*logsinx-1)`
`=>(dy)/(dx)=(1-(x+y)ycotx)/((x+y)logsinx-1)`
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