If `y=(sinx-cosx)^(sinx-cosx)` ,`pi/4 ltxlt (3pi)/ 4` find dy/dx

Explanation

`y=(sin x-cos x)^{sin x-cos x}`,
`pi/ 4ltxlt3pi / 4`

on taking log both sides, we the

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

`log y=(sin x-cos x) cdot log (sin x-cos x) `

`[ log m^{n}=n log m]`

On differentiating both sides w.r.t.` x`, we get

`frac{1}{y} cdot frac{d y}{d x}=(sin x-cos x) times frac{d}{d x} log (sin x-cos x)+log `

`(sin x-cos x) times frac{d}{d x}(sin x-cos x)`

[by using product rule of derivative]

or `frac{1}{y} frac{d y}{d x}=(sin x-cos x) cdot frac{1}{(sin x-cos x)} cdot frac{d}{d x}(sin x-cos x)+log (sin x-cos x) cdot(cos x+sin x)`

or `frac{1}{y} frac{d y}{d x}=(sin x-cos x) frac{1}{sin x-cos x} cdot frac{d}{d x}(sin x-cos x)`

`quad+log (sin x-cos x) frac{1}{sin x-cos x} `

`text { or } frac{1}{y} frac{d y}{d x}=(sin x-cos x) frac{1}{sin x-cos x}(cos x+sin x) +log (sin x-cos x) cdot(cos x+sin x)`

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

`text { or } quad[1+log (sin x-cos x)] `

`therefore frac{d y}{d x}=(sin x-cos x)^{(sin x-cos x)} cdot(cos x+sin x) `

`[1+log (sin x-cos x)]`