Question

How do you find the derivative of `y=x^lnx`?

Answer 1

By taking the natural logarithm of both sides:

`lny = ln(x^(lnx))`

Differentiate both sides:

`d/dx(lny) =d/dx(lnx(lnx))`

`1/y(dy/dx) = square`

Inset: `square`

We need to differentiate `lnx(lnx)`. By the product rule:

`[lnx(lnx)]' = 1/x xx lnx + 1/x xx lnx = lnx/x + lnx/x = (2lnx)/x`

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

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

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

Hopefully this helps!