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

Question

7023 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-27T17:07:25

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.

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