If y=tan ^{-1}(sin sqrt{x})+operatorname{cose | vedclass

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(B) Given $y=	an ^{-1}(\sin \sqrt{x})+\operatorname{cosec}^{-1}\left(e^{2 x+1}ight)$.Applying the chain rule for differentiation:$\frac{d}{dx}(	an

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$\frac{\cos \sqrt{x}}{2 \sqrt{x}\left(1+\sin ^2 \sqrt{x}\right)}-\frac{2}{\sqrt{e^{4 x+2}-1}}$

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(B) Given $y=	an ^{-1}(\sin \sqrt{x})+\operatorname{cosec}^{-1}\left(e^{2 x+1}ight)$.Applying the chain rule for differentiation:$\frac{d}{dx}(	an^{-1}(\sin \sqrt{x})) = \frac{1}{1+(\sin \sqrt{x})^2} \cdot \cos \sqrt{x} \cdot \frac{1}{2\sqrt{x}} = \frac{\cos \sqrt{x}}{2\sqrt{x}(1+\sin^2 \sqrt{x})}$.For the second term, using $\frac{d}{dx}(\operatorname{cosec}^{-1}(u)) = -\frac{1}{|u|\sqrt{u^2-1}} \cdot \frac{du}{dx}$:$\frac{d}{dx}(\operatorname{cosec}^{-1}(e^{2x+1})) = -\frac{1}{e^{2x+1}\sqrt{(e^{2x+1})^2-1}} \cdot e^{2x+1} \cdot 2 = -\frac{2}{\sqrt{e^{4x+2}-1}}$.Combining these, we get $\frac{dy}{dx} = \frac{\cos \sqrt{x}}{2 \sqrt{x}(1+\sin ^2 \sqrt{x})} - \frac{2}{\sqrt{e^{4 x+2}-1}}$.

If $y=\tan ^{-1}(\sin \sqrt{x})+\operatorname{cosec}^{-1}\left(e^{2 x+1}\right)$, then $\frac{d y}{d x}=$

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