Question

Can you differentiate x^x?

Answer 1

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

Explanation:

This question will require the knowledge of implicit differentiation and logarithms:

Let: `y = x^x`

Taking natural logs:

`lny = lnx^x`

Now using our knowledge of logs :

`log_gamma alpha^beta -= beta log_gamma alpha`

`=> lny = xlnx`

Implicitly differentiating, also using product rule

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

As `color(blue)(d/(dx) ( lnx ) = 1/x`

and `color(green)(d/(dx) xlnx = x* d/(dx) ( lnx) + lnx * d/(dx) ( x )`

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

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

Graphs:

`color(purple)(y= x^x`:

`color(orange)( y = d/(dx) ( x^x ) = x^x ( lnx + 1 ):`

Answer 2

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

Explanation:

You can also differentiate explicitly by writing the function as:

`x^x = (e^lnx)^x = e^(xlnx)`

so that:

`d/dx (x^x) = d/dx (e^(xlnx))`

and then using the chain rule:

`d/dx (x^x) = e^(xlnx) d/dx(xlnx) = x^x(1+lnx)`