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

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

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

$y = \log (\sec e^{x^2})$ so $\frac{dy}{dx} = \frac{1}{\sec e^{x^2}} \frac{d}{dx}(\sec e^{x^2})$ $= \frac{\sec e^{x^2} \tan e^{x^2}}{\sec e^{x^2}} \frac{d}{dx} (e^{x^2})$ $\frac{dy}{dx} = 2x e^{x^2} \tan e^{x^2}$