If `(sin"x")^y=x+y ,p rov et h a t(dy)/(dx)=(1-(x+y)ycotx)/((x+y)logsinx-1)`

Explanation

Given,`(sin"x")^y=x+y`

Taking log both sides,

=`ylog(sin x)=log(x +y) `

Differentiating it with respect to x using the chain rule and product rule,

=`(d/dx)(ylog(sin x))=(d/dx)(log(x +y)) `

=`y(d/dx)(log(sin x))+log(sin x)(dy/dx)=1/(x+y)(d/dx)(x +y) `

=`(ycosx)/(sinx)log(sin x)(dy/dx)=1/(x+y)+1/(x+y)(dy/dx)`

=`(dy/dx)(log(sin x)-1/(x+y))=1/(x+y)-ycotx`

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