Question

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

Answer 1

`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`