If y=sqrt{2 x+cos ^2left(2 x+frac{pi}{4}right | vedclass

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(A) Given,$y=\sqrt{2 x+\cos ^2\left(2 x+\frac{\pi}{4}\right)}$.Squaring both sides,we get $y^2=2 x+\cos ^2\left(2 x+\frac{\pi}{4}\right)$.Differentiat

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$\frac{2 \sqrt{2}}{\sqrt{\pi+1}}$

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(A) Given,$y=\sqrt{2 x+\cos ^2\left(2 x+\frac{\pi}{4}ight)}$.Squaring both sides,we get $y^2=2 x+\cos ^2\left(2 x+\frac{\pi}{4}ight)$.Differentiating with respect to $x$:$2 y \frac{d y}{d x} = 2 + 2 \cos \left(2 x+\frac{\pi}{4}ight) \cdot \left(-\sin \left(2 x+\frac{\pi}{4}ight)ight) \cdot 2$.Using the identity $\sin(2	heta) = 2\sin	heta \cos	heta$,we have:$2 y \frac{d y}{d x} = 2 - 2 \sin \left(4 x+\frac{\pi}{2}ight) = 2 - 2 \cos(4x)$.So,$\frac{d y}{d x} = \frac{1 - \cos(4x)}{y}$.At $x = \frac{\pi}{4}$,$y = \sqrt{2(\frac{\pi}{4}) + \cos^2(\frac{\pi}{2} + \frac{\pi}{4})} = \sqrt{\frac{\pi}{2} + \sin^2(\frac{\pi}{4})} = \sqrt{\frac{\pi}{2} + \frac{1}{2}} = \sqrt{\frac{\pi+1}{2}}$.Substituting these values:$\left. \frac{d y}{d x} ight|_{x=\frac{\pi}{4}} = \frac{1 - \cos(\pi)}{\sqrt{\frac{\pi+1}{2}}} = \frac{1 - (-1)}{\frac{\sqrt{\pi+1}}{\sqrt{2}}} = \frac{2 \sqrt{2}}{\sqrt{\pi+1}}$.

If $y=\sqrt{2 x+\cos ^2\left(2 x+\frac{\pi}{4}\right)}$,then find $\frac{d y}{d x}$ at $x=\frac{\pi}{4}$.

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