If y = sin (sqrt {sin x + cos x} ),then frac{ | vedclass

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(C) Given $y = \sin (\sqrt {\sin x + \cos x} )$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \cos (\sqrt {\sin x + \c

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$\frac{1}{2} \frac{\cos \sqrt {\sin x + \cos x}}{\sqrt {\sin x + \cos x}} (\cos x - \sin x)$

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(C) Given $y = \sin (\sqrt {\sin x + \cos x} )$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \cos (\sqrt {\sin x + \cos x}) \cdot \frac{d}{dx} (\sqrt {\sin x + \cos x})$Using the derivative of $\sqrt{u}$ which is $\frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}$:$\frac{dy}{dx} = \cos (\sqrt {\sin x + \cos x}) \cdot \frac{1}{2\sqrt{\sin x + \cos x}} \cdot \frac{d}{dx} (\sin x + \cos x)$Since $\frac{d}{dx} (\sin x + \cos x) = \cos x - \sin x$:$\frac{dy}{dx} = \frac{1}{2} \frac{\cos (\sqrt {\sin x + \cos x})}{\sqrt {\sin x + \cos x}} (\cos x - \sin x)$.

If $y = \sin (\sqrt {\sin x + \cos x} )$,then $\frac{dy}{dx} = $

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