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

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Aicharya · Accepted Answer

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 $\begin{aligned} 2 y \frac{d y}{d x} & =2-2 \sin \left(4 x+\frac{\pi}{2}\right) \ & =2-2 \cos 4 x \Rightarrow \frac{d y}{d x}=\frac{1-\cos 4 x}{y} \ \left.\Rightarrow \quad \frac{d y}{d x}\right|_{x=\frac{\pi}{4}} & =\frac{1-(-1)}{\sqrt{\frac{\pi}{2}+\frac{1}{2}}}=\frac{2 \sqrt{2}}{\sqrt{\pi+1}} \end{aligned}$

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

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