# What is the derivative of e^((lnx)^2)?

Apr 4, 2015

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

Derivate will be $u ' \cdot {e}^{u}$

So :

$\left({\left(\ln \left(x\right)\right)}^{2}\right) ' = \left({u}^{n}\right) ' = n \cdot u ' {u}^{n - 1}$ here $n = 2$

$\left({\left(\ln \left(x\right)\right)}^{2}\right) ' = 2 \cdot \frac{1}{x} \cdot \ln \left(x\right) = \frac{2 \ln \left(x\right)}{x}$

Finally we have (e^((ln(x))^2) )' = (2ln(x))/x*e^((ln(x))^2