Question

What is the derivative of x to the x? d/dx (x^x)

If you answer, thank you! I always see this problem but I don't know how to answer it.

Answer 1

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

Explanation:

`y = x^x`

`Lny =xlnx`

Apply implicit differentiation, standard differential and the product rule.

`1/y* dy/dx = x*1/x +lnx*1`

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

Substitute `y = x^x`

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

Answer 2

`(x^x)(ln(x) + 1)`

Explanation:

`dy/dx [x^x] = dy/dx [e^{xln(x)}]`
Let `u = xln(x)` and thus, `x^x = e^u`

Apply chain rule:
`dy/dx = dy/du * du/dx`
`= d/du [ e^u ] * d/dx [xln(x) ]`

Derivative of `e^u` is itself, Derivative of `ln(x)` is `\frac{1}{x}` and also apply product rule `d/dx [f(x)g(x)] = f'(x)g(x) + g'(x)f(x)`

`= (e^u) [(x)(1/x) + (1)(ln(x))]`
`= (x^x) [(x)(1/x) + (1)(ln(x))]`
`= (x^x) [1 +ln(x)]`