.or y= lnx^(tanx).

So: with logarithmic differentiation:

lny= ln(lnx^(tanx))

1/y dy/dx = tanx ln (lnx)

1/y dy/dx = sec²x ln(lnx) + tanx(ln(1/x)

1/y dy/dx= sec²xln(lnx) + tanx(???)

My Question Is: What would be the derivative of ln(1/x)

The final answer is: dy/dx= (lnx)^tanx * sec²ln(lnx) + tanx/(xlnx)

How is the derivative of ln(1/x)= xlnx?

It isn't:

Tonio

Thank you.