1. ## Derivative of y=ln(lnx)

Find the equation of the tangent line to the curve y=ln(lnx) which is perpendicular to the line ex+y =0

So I took the derivative and got 1/(xlnx)
I know that the slope of the normal line is 1/e so i substituted that in for y' and tried solving for x. this is where I was stuck
1/e = 1/(xlnx)
xlnx = e

2. Originally Posted by john-1
Find the equation of the tangent line to the curve y=ln(lnx) which is perpendicular to the line ex+y =0

So I took the derivative and got 1/(xlnx)
I know that the slope of the normal line is 1/e so i substituted that in for y' and tried solving for x. this is where I was stuck
1/e = 1/(xlnx)
xlnx = e
Looks correct to me.

$x \ln x = e \implies \ln x^x = e \implies x^x = e^e$

Therefore $x=e$.

It's sort of a weird way to solve for $x$ but it gives the correct solution.