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Find (dy)/(dx), y=(sinx)^(cosx)+(cosx)^...

Find `(dy)/(dx),` `y=(sinx)^(cosx)+(cosx)^(sinx)`

Text Solution

Verified by Experts

Given that,
`y=(sinx)^(cosx)+(cosx)^(sinx)`
Taking log both side and we get,
`logy=log(sinx)^(cosx)+log(cosx)^(sinx)`
`=>log y=cosx*log sinx + sinx *log cos x`
On differentiating with respect to x and we get,
`d/(dx) logy=cosx d/(dx) log sinx+log sinx d/(dx) cosx+sinx d/(dx) log cosx+log cos xd/(dx)sinx`
`=>1/y(dy)/(dx)=cosx/sinx(cosx)+log sinx(-sinx)+sinx/cosx(-sinx)+logcosx cosx`
`=>1/y(dy)/(dx)=(cos^2x)/sinx -(sin^2x)/cosx-sinx log sinx+cosx log cosx`
`=>1/y(dy)/(dx)=(cos^2x)/sinx -(sin^2x)/cosx+cosx log cosx+sinx log sinx`
`=>1/y (dy)/(dx)=(cos^3x-sin^3x)/(sinx*cosx)+log*((cosx)^(cosx))/((sinx)^(sinx))`
`=>(dy)/(dx)=y[(cos^3x-sin^3x)/(sinx*cosx)+log*((cosx)^(cosx))/((sinx)^(sinx))]`
`=>(dy)/(dx)=(sinx)^(cosx)+(cosx)^(sinx)[(cos^3x-sin^3x)/(sinx*cosx)+log*((cosx)^(cosx))/((sinx)^(sinx))]`
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