# f: R-> C = x^x if x>0 and f(x):=0 if x<=0 we must show that f differentiable in 0 if Re(z) >1The derivative of f(x) is equal to df/dx = x^x * ( ln(x) +1 )?

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### 2 Answers

You need to remember the formula of derivative of complex function: first you differentiate `x^x` as it would be a power function, then you need to differentiate `x^x` as it would be an exponential function.

`(x^x)' = x*(x^(x-1))*(x)' + (x^x)*(ln x)*(x)'`

`(x^x)' = (x^(x-1+1))*(1) + (x^x)*(ln x)*(1)`

`(x^x)' =x^x + (x^x)*(ln x)`

You need to factor out `x^x` such that:

`(x^x)' = (x^x)*(1 + ln x)`

If f(x) differentiable at x = 0, then the left hand side derivative of the function at x=0 is equal to the right hand side derivative at x=0.

`f'(0) = lim_(x-gt0) ((x^x)*(1 + ln x))/(x) = lim_(x-gt0) (1 + ln x)/(x/(x^x)) = lim_(x-gt0) (1 + ln x)/ (x^(1-x)) = 00`

The right derivative is derivative of constant function => f'(0) = 0.

**Since the left hand side derivative of the function at x=0 is not equal to the right hand side derivative at x=0, hence, the function is not differentiable at x=0.**

Thanks for the answer, but this can not yet be the final story. The exercise we have to complete is asking us to show that f is differentiable even though, but at the condition that Re z >1.