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

Apr 27, 2015

Use $\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \frac{\mathrm{du}}{\mathrm{dx}}$ and the power and chain rule:

$y = {e}^{{\left(\ln x\right)}^{2}}$

$y ' = {e}^{{\left(\ln x\right)}^{2}} \cdot \frac{d}{\mathrm{dx}} \left({\left(\ln x\right)}^{2}\right) = {e}^{{\left(\ln x\right)}^{2}} \cdot 2 \left(\ln x\right) \left(\frac{1}{x}\right)$

$y ' = \frac{2 {e}^{{\left(\ln x\right)}^{2}} \ln x}{x}$

Note on rewriting

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

$y = {e}^{{\left(\ln x\right)}^{2}} = {\left({e}^{\ln} x\right)}^{\ln} x = {x}^{\ln} x$

And $y = {x}^{\ln} x$ can be differentiated by logarithmic differentiation, but it doesn't seem easier than the approach taken above.