If $y = \log_e(\sec e^{x^2})$, then $\frac{dy}{dx} =$

$2xe^{x^2}\tan e^{x^2}$

The correct answer is Option (2) → $2xe^{x^2}\tan e^{x^2}$

$y=\ln\!\left(\sec\!\left(e^{x^{2}}\right)\right)$

$\frac{dy}{dx}=\frac{d}{dx}\big[\ln(\sec u)\big]\Big|_{u=e^{x^{2}}} =\tan(u)\cdot\frac{du}{dx} =\tan\!\left(e^{x^{2}}\right)\cdot\frac{d}{dx}\!\left(e^{x^{2}}\right)$

$\frac{d}{dx}\!\left(e^{x^{2}}\right)=e^{x^{2}}\cdot 2x$

$\Rightarrow\ \frac{dy}{dx}=2x\,e^{x^{2}}\tan\!\left(e^{x^{2}}\right)$

$2x\,e^{x^{2}}\tan\!\left(e^{x^{2}}\right)$