`x(cosx)^(x)+(x sinx)^((1)/(x))`

Explanation

माना `y=(x cos x)^(x)+(x sin x)^(1//x)`
माना `u=(x cos x)^(x)` तथा `v=(x sin x)^(1//x)`
`therefore" "y=u+v`
`rArr" "(dy)/(dx)=(du)/(dx)+(dv)/(dx)`
अब `u=(x cosx)^(x)`
`rArr" "logu=log(xcos x)^(x)=x (logx +log cosx)`
`rArr" "(1)/(u)(du)/(dx)=x.(d)/(dx)(logx+logcos x)+(log x+log cosx)(d)/(dx)x`
`rArr" "(du)/(dx)=u[x((1)/(x)-(sinx)/(cosx))+(logx+logcos x).1]`
`rArr" "(du)/(dx)=(x cos x)^(x)[1- x tanx +logx+log cos x]` तथा `" "v=(x sin x)^(1//x)`
`rArr" "logv=log(x sin x)^(1//x)=(1)/(x)(logx+logsinx)`
`rArr" "(1)/(v)(dv)/(dx)=(1)/(x)(d)/(dx)(logx+logsin x)+(logx+log sinx)(d)/(dx)((1)/(x))`
`rArr" "(dv)/(dx)v[(1)/(x)((1)/(x)+(cosx)/(sinx))-(1)/(x^(2))(logx+log sin x)]`
`rArr" "(dv)/(dx)=(x sinx)^(1//x)[(1+x cot x-log (x sinx))/(x^(2))]`
समीकरण (1 ) से
`(dy)/(dx)=(x cos x)^(x)[1-k x tanx +log x+log cosx]+(x sin x)^(1/x)[1-x tan x+log(x sin x)]`