Put $\mathit{x}=\cos 2\mathit{\theta }$
$\Rightarrow \mathrm{y}=\mathit{s}\mathit{i}\mathit{n}{}^{-1}$
$\left(\frac{\sqrt{1+\cos 2\mathit{\theta }}}{2}+\frac{\sqrt{1-\cos 2\mathit{\theta }}}{2}\right)$
$\Rightarrow \mathrm{y}=\mathit{s}\mathit{i}\mathit{n}{}^{-1}$
$\left(\frac{\sqrt{2{\cos }^{2}2\mathit{\theta }}}{2}+\frac{\sqrt{2\cdot \mathit{s}\mathit{i}\mathit{n}{}^2\mathit{\theta }}}{2}\right)$
$\Rightarrow \mathit{y}=\mathit{s}\mathit{i}\mathit{n}{}^{-1}\left(\frac{\cos 2\mathit{\theta }}{\sqrt{2}}+\frac{\mathit{s}\mathit{i}\mathit{n}2\mathit{\theta }}{\sqrt{2}}\right)$
$\Rightarrow \mathit{y}=\mathit{s}\mathit{i}\mathit{n}{}^{-1}(\mathit{s}\mathit{i}\mathit{n}\left(\frac{\mathit{\pi }}{4}+2\mathit{\theta }\right).$

$\Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{\theta }}=2$

$\Rightarrow \frac{\mathit{d}\mathit{\theta }}{\mathit{d}\mathit{x}}=\frac{-1}{4\sqrt{1-{\mathit{x}}^2}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{2\sqrt{1-{\mathrm{x}}^2}}$

OR

As we know that exponential and cosine functions are continuous and differentiable on $\mathrm{R}$ .
Let us find the values of the function at an extreme
$\Rightarrow \mathrm{f}\left(-\frac{\mathit{\pi }}{2}\right)={\mathrm{e}}^{-\frac{\mathit{\pi }}{2}}\cos \left(-\frac{\mathit{\pi }}{2}\right)$
$\Rightarrow \mathrm{f}\left(-\frac{\mathit{\pi }}{2}\right)={\mathrm{e}}^{-\frac{\mathit{\pi }}{2}}\times 0$
$\Rightarrow \mathrm{f}\left(-\frac{\mathit{\pi }}{2}\right)=0$
$\Rightarrow \mathrm{f}\left(\frac{\mathit{\pi }}{2}\right)={\mathrm{e}}^{\frac{\mathit{\pi }}{2}}\cos \left(\frac{\mathit{\pi }}{2}\right)$
$\Rightarrow \mathrm{f}\left(\mathit{\pi }\right)={\mathrm{e}}^{\frac{\mathit{\pi }}{2}}\times 0$
$\Rightarrow \mathrm{f}\left(\mathit{\pi }\right)=0$
Here, ${\mathit{f}}^{\prime }\left(-\mathit{\pi }∕2\right)=\mathit{f}\left(\mathit{\pi }∕2\right)$ , therefore there exist a $\mathit{c}\in \left(-\mathit{\pi }∕2,\mathit{\pi }∕2\right)$ such that ${\mathit{f}}^{\prime }\left(\mathit{c}\right)=0$ .
Let us find the derivative of $\mathit{f}\left(\mathit{x}\right)$
$\Rightarrow {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\frac{\mathrm{d}\left({\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x}\right)}{\mathrm{dx}}$
$\Rightarrow {\mathit{f}}^{\prime }\left(\mathit{x}\right)=\cos \mathit{x}\frac{\mathit{d}\left({\mathit{e}}^{\mathit{x}}\right)}{\mathit{d}\mathit{x}}+{\mathit{e}}^{\mathit{x}}\frac{\mathit{d}\left(\cos \mathit{x}\right)}{\mathit{d}\mathit{x}}$
$\Rightarrow {\mathit{f}}^{\prime }\left(\mathit{x}\right)={\mathit{e}}^{\mathit{x}}\left(-\mathit{s}\mathit{i}\mathit{n}\mathit{x}+\cos \mathit{x}\right)$

$\Rightarrow {\mathit{e}}^{\mathit{c}}\left(-\mathit{s}\mathit{i}\mathit{n}\mathit{c}+\cos \mathit{c}\right)=0$
$\Rightarrow -\mathit{s}\mathit{i}\mathit{n}\mathit{c}+\cos \mathit{c}=0$
$\Rightarrow \frac{-1}{\sqrt{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{c}+\frac{1}{\sqrt{2}}\cos \mathit{c}=0$
$\Rightarrow -\mathit{s}\mathit{i}\mathit{n}\left(\frac{\mathit{\pi }}{4}\right)\mathrm{}\mathit{s}\mathit{i}\mathit{n}\mathit{c}+\cos \left(\frac{\mathit{\pi }}{4}\right)\cos \mathit{c}=0$
$\Rightarrow \cos \left(\mathit{c}+\frac{\mathit{\pi }}{4}\right)=0$
$\Rightarrow \mathit{c}+\frac{\mathit{\pi }}{4}=\frac{\mathit{\pi }}{2}$
$\Rightarrow \mathit{c}=\frac{\mathit{\pi }}{4}\mathit{E}\left(-\frac{\mathit{\pi }}{2},\frac{\mathit{\pi }}{2}\right)$
Thus, Rolle's theorem is verified.