Differentiate `x^(xcosx)+(x^2+1)/(x^2-1)` with respect to `x` :

Explanation

`"Let "y = x^(x"cos"x) + (x^(2) + 1)/(x^(2)-1)`
`"Let "u = x^(x"cos"x) " and "v=(x^(2)+1)/(x^(2)-1)`
`therefore y=u+v`
`rArr (dy)/(dx) = (du)/(dx) + (dv)/(dx) " ".....(1)`
`"Now", u=x^(x"cos"x)`
`rArr "log" u = "log"(x^(x"cos"x)) = x"cos"x * "log"x`
`rArr (1)/(u)(du)/(dx) = x "cos"x * (d)/(dx)"log"x + x"log"x * (d)/(dx)"cos"x + "cos"x * "log"x * (d)/(dx)x`
`rArr (du)/(dx) = u(x"cos"x * (1)/(x)-x"log"x * "sin"x + "cos"x * "log"x)`
`= x^(x"cos"x) ("cos"x-x"log"x"sin"x + "cos"x"log"x)`
`"and "v = (x^(2) + 1)/(x^(2)-1)`
`rArr (dv)/(dx) = ((x^(2)-1)(d)/(dx)(x^(2)+1) - (x^(2)+1)(d)/(dx)(x^(2)-1))/(x^(2)-1)^(2)`
`= ((x^(2)-1) * 2x - (x^(2)+1) * 2x)/(x^(2)-1)^(2)`
`=-(4x)/(x^(2)-1)^(2)`
From equation (1)
`(dy)/(dx) = x^(x"cos"x)("cos" x - x"log"x"sin"x + "cos"x"log"x)-(4x)/(x^(2)-1)^(2)`