Put $x = \cos 2\theta$ $\Rightarrow y = \sin^{-1}$ $\left( \frac{\sqrt{1+\cos 2\theta}}{2} + \frac{\sqrt{1-\cos 2\theta}}{2} \right)$ $\Rightarrow y = \sin^{-1}$ $\left( \frac{\sqrt{2 \cos^2 2\theta}}{2} + \frac{\sqrt{2 \cdot \sin^2 \theta}}{2} \right)$ $\Rightarrow y = \sin^{-1} \left( \frac{\cos 2\theta}{\sqrt{2}} + \frac{\sin 2\theta}{\sqrt{2}} \right)$ $\Rightarrow y = \sin^{-1} ( \sin \left( \frac{\pi}{4} + 2\theta \right) .$ $\Rightarrow y = \frac{\pi}{4} + 2\theta \; \text{. } \;$ $\Rightarrow \frac{dy}{d\theta} = 2$ $\;\text{Put}\; \theta = \frac{\cos^{-1} x}{2}$ $\Rightarrow \frac{d\theta}{dx} = \frac{-1}{4 \sqrt{1-x^2}}$ $\therefore \frac{dy}{dx} = \frac{-1}{2 \sqrt{1-x^2}}$ OR As we know that exponential and cosine functions are continuous and differentiable on $R$ .Let us find the values of the function at an extreme $\Rightarrow f\left(-\frac{\pi}{2}\right) = e^{-\frac{\pi}{2}} \cos\left(-\frac{\pi}{2}\right)$ $\Rightarrow f\left(-\frac{\pi}{2}\right) = e^{-\frac{\pi}{2}} \times 0$ $\Rightarrow f\left(-\frac{\pi}{2}\right) = 0$ $\Rightarrow f\left(\frac{\pi}{2}\right) = e^{\frac{\pi}{2}} \cos\left(\frac{\pi}{2}\right)$ $\Rightarrow f(\pi) = e^{\frac{\pi}{2}} \times 0$ $\Rightarrow f(\pi) = 0$Here, $f'\left(-\frac{\pi}{2}\right) = f\left(\frac{\pi}{2}\right)$ , therefore there exist a $c \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $f'(c)=0$ .Let us find the derivative of $f(x)$ $\Rightarrow f'(x) = \frac{d \left( e^{x} \cos x \right)}{dx}$ $\Rightarrow f'(x) = \cos x \frac{d (e^x)}{dx} + e^x \frac{d(\cos x)}{dx}$$\Rightarrow f'(x) = e^x ( -\sin x + \cos x )$$\;\text{Here,}\; f'(c)=0$$\Rightarrow e^c ( -\sin c + \cos c ) = 0$$\Rightarrow -\sin c + \cos c = 0$$\Rightarrow \frac{-1}{\sqrt{2}} \sin c + \frac{1}{\sqrt{2}} \cos c = 0$$\Rightarrow -\sin \left( \frac{\pi}{4} \right) \sin c + \cos \left( \frac{\pi}{4} \right) \cos c = 0$$\Rightarrow \cos \left( c + \frac{\pi}{4} \right) = 0$$\Rightarrow c + \frac{\pi}{4} = \frac{\pi}{2}$$\Rightarrow c = \frac{\pi}{4} E \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$Thus, Rolle's theorem is verified.