Question

How do you find the first and second derivative of `(lnx)^(x)`?

Answer 1

Let:

`y=(lnx)^x`

The easiest way to differentiate this is to take the natural logarithm of both sides.

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

Simplify using the logarithm rule:

`ln(y)=xln(lnx)`

Differentiate both sides. The left-hand side will need chain rule, similar to implicit differentiation, and the right-hand side will need chain rule and product rule.

`1/y*dy/dx=d/dx(x)*ln(lnx)+x*d/dxln(lnx)`

`1/y*dy/dx=ln(lnx)+x*1/lnx*d/dxlnx`

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

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

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

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

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

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

The steps to find the second derivative are far too lengthy, and the work is rather pointless, but if you wish try to find it, the second derivative is:

`(d^2y)/(dx^2)=((lnx)^(x-2)(lnx(xln(lnx)(lnx*ln(lnx)+2)+1)+x-1))/x`