Question

`d/dx (x^x)?`

Answer 1

`(dy)/(dx)=x^x(1+lnx)`

Explanation:

we can use logarithmic differentiation

`d/(dx)(x^x)`

let

`y=x^x`

take natural logs of both sides

`lny=xlnx`

we now differentiate `wrt" "x`

the '`LHS` will be need the chain rule, the `RHS` the product rule

`d/(dx)(lny)=d/(dx)(xlnx)`

`1/y(dy)/(dx)=lnxd/(dx)(x)+xd/(dx)(lnx)`

`1/y(dy)/(dx)=lnx+xxx1/x`

`(dy)/(dx)=y(1+lnx)`

`:.(dy)/(dx)=x^x(1+lnx)`