If `y=x^x` , find `(dy)/(dx)` at `x=e` .

Explanation

Given, `y=x^(x)` , taking log on both sides
` therefore log y = log (x^(x))`
`rArr log y = x log x `
Differentiating w.r.t.x., by using rule
` rArr (1)/(y) (dy)/(dx) = x xx(1)/(x) + 1 xx logx `
` rArr (1)/(y) . (dy)/(dx) = 1 + log x `
` rArr (dy)/(dx) = y ( 1 + log x)`
` rArr (dy)/(dx) = x^(x) (1 + log x)`