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

1 Answer
Apr 27, 2015

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.