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

Given: $y = \log_e(\sec e^{x^2})$

Since $\log_e = \ln$ and $\sec \theta = \frac{1}{\cos \theta}$:

$y = \ln\left(\frac{1}{\cos e^{x^2}}\right)$

Using the property $\ln\left(\frac{1}{a}\right) = -\ln(a)$:

$y = -\ln(\cos e^{x^2})$

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Differentiating with respect to $x$, where $u = \cos e^{x^2}$:

$\frac{dy}{dx} = -\frac{1}{\cos e^{x^2}} \cdot \frac{d}{dx}(\cos e^{x^2})$

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For $\frac{d}{dx}(\cos e^{x^2})$, let $v = e^{x^2}$:

$\frac{d}{dx}(\cos e^{x^2}) = -\sin e^{x^2} \cdot \frac{d}{dx}(e^{x^2})$

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For $\frac{d}{dx}(e^{x^2})$, let $w = x^2$:

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

$\frac{d}{dx}(e^{x^2}) = 2xe^{x^2}$

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Combining the results:

$\frac{d}{dx}(\cos e^{x^2}) = -\sin e^{x^2} \cdot 2xe^{x^2}$

$\frac{dy}{dx} = -\frac{1}{\cos e^{x^2}} \cdot (-\sin e^{x^2} \cdot 2xe^{x^2})$

$\frac{dy}{dx} = \frac{\sin e^{x^2}}{\cos e^{x^2}} \cdot 2xe^{x^2}$

Using $\frac{\sin \theta}{\cos \theta} = \tan \theta$:

$\frac{dy}{dx} = 2xe^{x^2} \tan e^{x^2}$