Mathematics › Differentiation › Differentiation of composite functions
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…
Find the value of $k$ if $\frac{d}{d x}\left\{\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos (4 x)}}}}\right\}$ $=k \sec \left(\frac{x}{2}\right) \tan \left(\frac{x}{2}\right)$
$\frac{1}{2}$ 2 1 $\frac{1}{8}$ Solution
$\begin{aligned} & \text { Let } y=\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4 x}}}} \\ & \because \quad 1+\cos 2 A=2 \cos ^2 A \\ & \therefore y=\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2(1+\cos 4 x)}}}} \\ & =\frac{2}{\sqrt{2+\sqrt{2+\sqrt{4 \cos ^2 2 x}}}}=\frac{2}{\sqrt{2+\sqrt{2+2 \cos 2 x}}} \\ & =\frac{2}{\sqrt{2+\sqrt{2(1+\cos 2 x)}}}=\frac{2}{\sqrt{2+\sqrt{4 \cos ^2 x}}} \\ & =\frac{2}{\sqrt{2+2 \cos x}}=\frac{2}{\sqrt{2(1+\cos x)}} \\ & y=\frac{2}{\sqrt{4 \cos ^2 \frac{x}{2}}} \\ & \end{aligned}$
$
\begin{gathered}
\Rightarrow y=\frac{2}{2 \cos \frac{x}{2}} \\
y=\sec \frac{x}{2}
\end{gathered}
$
Differentiate w.r.t. $x$,
$
\frac{d y}{d x}=\sec \frac{x}{2} \tan \frac{x}{2}\left(\frac{1}{2}\right) \text { [using chain rule] }
$
Now, according to the question,
$
\begin{aligned}
\frac{d y}{d x} & =k \sec \frac{x}{2} \tan \frac{x}{2} \\
\sec \frac{x}{2} \tan \frac{x}{2} \cdot \frac{1}{2} & =k \sec \frac{x}{2} \tan \frac{x}{2} \\
\therefore \quad k & =1 / 2
\end{aligned}
$
Asked in: AP EAMCET 2021 (24 Aug Shift 1)
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