Question

How do you find the derivative of `y = x^(ln x)`?

Answer 1

Take the natural logarithm of both sides:

`lny = ln(x^lnx)`

`lny = lnx(lnx)`

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

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

`1/y(dy/dx) = (2lnx)/x`

`dy/dx= ((2lnx)/x)/(1/y)`

`dy/dx= y xx (2lnx)/x`

`dy/dx = x^lnx xx (2lnx)/x`

`dy/dx = x^(lnx - 1) xx 2lnx`

Hopefully this helps!