derivative of $e^{\ln x^2}-3x^7$

Question

$$e^{\ln x^2}-3x^7$$

The first term: $=e^v$
$v=\ln x^2=u^2$

$v\;'=2uu\;'=(2\ln x)\dfrac{1}{x}=\dfrac{2\ln x}{x}$

$\dfrac{e^{\ln x^2}2\ln x}{x} +21x^{-8}$

How do I simplify further? I don't understand why the first term turns out to be $2x$

Answer 1

Remember that $$e^{\ln a}=a$$ Now the derivative should be trivial.

Answer 2

$v=\ln x^2=u^2$ ?? it would be right if $(ln x)^2$

it should be $v=\ln x^2=2u$