Originally Posted by
Mush Yes. As I said before $\displaystyle 2x + 3 \ln(x) $ is NOT the derivative of $\displaystyle 3x. \ln {x} $
You have carried out your differentiation incorrectly on the RHS of the equation AFTER you've taken the natural logarithm of both sides.
$\displaystyle y = x^{3x} $
$\displaystyle \ln(y) = \ln(x^{3x}) $
$\displaystyle \ln(y) = 3x \ln (x) $
$\displaystyle \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (3x . \ln(x)) $
You must use the product rule on the RHS to differentiate.