Given, \(y=\sqrt{2 x+\cos ^2\left(2 x+\frac{\pi}{4}\right)}\)
So, at \(x=\frac{\pi}{4}\)
\(\begin{aligned}
y & =\sqrt{\frac{\pi}{2}+\cos ^2\left(\frac{\pi}{2}+\frac{\pi}{4}\right)} \\
& =\sqrt{\frac{\pi}{2}+\sin ^2 \frac{\pi}{4}}=\sqrt{\frac{\pi}{2}+\frac{1}{2}}
\end{aligned}\)
Now, as \(y^2=2 x+\cos ^2\left(2 x+\frac{\pi}{4}\right)\)
On, differentiating w.r.t. \(x\), we get
\(\begin{aligned}
2 y \frac{d y}{d x} & =2-2 \sin \left(4 x+\frac{\pi}{2}\right) \\
& =2-2 \cos 4 x \Rightarrow \frac{d y}{d x}=\frac{1-\cos 4 x}{y} \\
\left.\Rightarrow \quad \frac{d y}{d x}\right|_{x=\frac{\pi}{4}} & =\frac{1-(-1)}{\sqrt{\frac{\pi}{2}+\frac{1}{2}}}=\frac{2 \sqrt{2}}{\sqrt{\pi+1}}
\end{aligned}\)