Find the derivative of `x^x-2^(sinx)` w.r.t. `x`

Explanation

`"Let" y = x^(x) - 2^("sin"x)`
`"Let " u=x^(x) and v=2^("sin"x)`
`therefore y=u-v rArr (dy)/(dx)= (du)/(dx)-(dv)/(dx) " "....(1)`
`"Now", u=x^(x)`
`rArr "log " u = "log " (x^(x)) = x" log " x`
Differentiate both sides w.r.t.x
`(1)/(u) (du)/(dx) = x*(d)/(dx)"log"x + "log" x * (d)/(dx)x`
`rArr (du)/(dx) = u[x*(1)/(x) + "log"x*1] = x^(x)(1+"log"x)`
`"and "v=2^("sin"x)`
`rArr (dv)/(dx) = (d)/(dx)2^("sin"x)`
`=2^("sin"x) " log" 2 * (d)/(dx) sinx`
`=2^("sin"x) * "log"2 * "cos"x`
Now from equation (1)
`(dy)/(dx) = x^(x) (1+ "log"x)- 2^("sin"x) * "log"2 * "cos"x`