Question

Write an equation of the tangent line to the curve given by y=(lnx)^lnx, at x=e.

Answer 1

let u = ln(x).  Then for y = ln(x)^ln(x) we have y=u^u and the chain rule gives us

dy/dx=dy/du*du/dx

Assume u>0 (there are problems otherwise) then ln(y) = ln(u^u) =ln(u)*u

Differentiate both sides using chain rule on the left and product rule on the right

dy/du*1/y = ln(u)+1/u*u=ln(u)+1

or dy/du = (ln(u)+1)*y = (ln(u) + 1)*u^u = (ln(ln(x))+1)*ln(x)^ln(x)

du/dx = d/du(ln(x)) = 1/x so the derivative is ((ln(ln(x))+1)*ln(x)^ln(x))/x

The slope of the tangent is (ln(ln(e))+1)*ln(e)^(ln(e))/e = (ln(1)+1)(1^1)/e=1/e

now y = 1 at x=e so y=mx+b is y=1/e(x)+1