Find the value of k if frac{d}{d x}left{frac{ | vedclass

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(A) Let $y = \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4 x}}}}$.Using the identity $1+\cos 2A = 2 \cos^2 A$,we simplify the expression step by step:$y =

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$\frac{1}{2}$

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(A) Let $y = \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4 x}}}}$.Using the identity $1+\cos 2A = 2 \cos^2 A$,we simplify the expression step by step:$y = \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2(1+\cos 4 x)}}}} = \frac{2}{\sqrt{2+\sqrt{2+\sqrt{4 \cos^2 2 x}}}} = \frac{2}{\sqrt{2+\sqrt{2+2 \cos 2 x}}}$.Continuing the simplification:$y = \frac{2}{\sqrt{2+\sqrt{2(1+\cos 2 x)}}} = \frac{2}{\sqrt{2+\sqrt{4 \cos^2 x}}} = \frac{2}{\sqrt{2+2 \cos x}} = \frac{2}{\sqrt{2(1+\cos x)}} = \frac{2}{\sqrt{4 \cos^2 \frac{x}{2}}}$.Thus,$y = \frac{2}{2 \cos \frac{x}{2}} = \sec \frac{x}{2}$.Differentiating with respect to $x$ using the chain rule:$\frac{dy}{dx} = \sec \frac{x}{2} 	an \frac{x}{2} \cdot \frac{1}{2}$.Comparing this with $k \sec \frac{x}{2} 	an \frac{x}{2}$,we get $k = \frac{1}{2}$.

Find the value of $k$ if $\frac{d}{d x}\left\{\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos (4 x)}}}}\right\} = k \sec \left(\frac{x}{2}\right) \tan \left(\frac{x}{2}\right)$

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