Question

How do you differentiate `y=e^((lnx)^2)`?

Answer 1

Use `d/dx(e^u) = e^u (du)/dx` and the power and chain rule:

`y=e^((lnx)^2)`

`y' = e^((lnx)^2) * d/dx((lnx)^2) = e^((lnx)^2)*2(lnx)(1/x)`

`y' = (2e^((lnx)^2)lnx)/x`

Note on rewriting

Trying to rewrite is helpful in many problems. It is less helpful in this case than one might hope:

`y = e^((lnx)^2)= (e^lnx)^lnx = x^lnx`

And `y=x^lnx` can be differentiated by logarithmic differentiation, but it doesn't seem easier than the approach taken above.