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

Explanation

We have, `(sin x)^{y}=(cos y)^{x}`

Taking log on both sides,

`log (sin x)^{y}=log (cos y)^{x} `

`Rightarrow y log (sin x)=x log (cos y)`

Differentiating with respect to `x`,

`frac{d}{d x}[y log sin x]=frac{d}{d x}[x log cos y] `

`Rightarrow y frac{d}{d x}(log sin x)+log sin x frac{d y}{d x}=x frac{d y}{d x}(log cos y)+log cos y frac{d}{d x}(x) `

`Rightarrow y(frac{1}{sin x}) frac{d}{d x}(sin x)+log sin x frac{d y}{d x}=frac{x}{cos y} frac{d}{d x}(cos y)+log cos y(1) `

`Rightarrow frac{y}{sin x}(cos x)+log sin x frac{d y}{d x}=frac{x}{cos y}(-sin y) frac{d y}{d x}+log cos y `

`Rightarrow y cot x+log sin x frac{d y}{d x}=-x tan y frac{d y}{d x}+log cos y `

`Rightarrow frac{d y}{d x}(log sin x+x tan y)=log cos y-y cot x`

`Rightarrow frac{d y}{d x}=frac{log cos y-y cot x}{log sin x+x tan y}`