1.

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

Answer» `"Let " y = (x+(1)/(x))^(x) +x^((1+(1)/(x)))`
`"Let " u =(x+(1)/(x))^(x) " and "v = x^(1+(1)/(x))`
`therefore y=u + v rArr (dy)/(dx) = (du)/(dx) + (dv)/(dx) " " ...(1)`
`"Now", u = (x+(1)/(x))^(x)`
`rArr "log " u = "log" (x+(1)/(x))^(x) = x" log"(x+(1)/(x))`
`rArr (1)/(u) (du)/(dx) = x* (d)/(dx) "log" (x+(1)/(x)) + "log"(x+(1)/(x))(d)/(dx) x`
`rArr (du)/(dx) = u [(x)/(x+(1)/(x))(1-(1)/x^(2))+"log" (x+(1)/(x))]`
`=(x + (1)/(x))^(x) [(x^(2)-1)/(x^(2)+1) + "log" (x+(1)/(x))]`
` "and "v = x^((1+(1)/(x))`
`rArr "log " v = "log" {x^((1+(1)/(x)))} = (1+(1)/(x))"log"x`
`rArr (1)/(v) (dv)/(dx) = (1+(1)/(x))(d)/(dx) "log"x + "log"x (d)/(dx) (1+(1)/(x))`
`rArr (dv)/(dx) = v[(1+(1)/(x))*(1)/(x)+"log"x (-(1)/(x^(2)))]`
`rArr (dv)/(dx) = x^((1+(1)/(x)))*(1)/(x^(2))[x+1-"log"x]`
`therefore "From equation (1)"`
`(dy)/(dx) = (x+(1)/(x))^(x)[(x^(2)-1)/(x^(2)+1)+ "log" (x+(1)/(x))] + x^((1+(1)/(x))) * (1)/(x^(2))[x+1-"log"x]`


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