What is the derivative of $e^{{(ln x)}^{2}}$ $e^((lnx)^2)$ ?

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Answer 1 · 2015-04-04T22:39:26

$u = {(ln (x))}^{2}$ $u = (ln(x))^2$

Derivate will be $u'*e^u$

So :

$((ln(x))^2)' = (u^n)' = n*u'u^(n-1)$ here $n = 2$

$((ln(x))^2)' = 2*1/x*ln(x) = (2ln(x))/x$

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

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