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Answer: $\frac{\mathrm{d} y}{\mathrm{~d} \mathrm{x}}=-1$

Hint:

$\frac{\mathrm{d}}{\mathrm{dx}} \text { (constants) }=0$

$\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$

Given:

$\cos ^{-1}\left\{\frac{\cos x+\sin x}{\sqrt{2}}\right\}$

$-\frac{\mathrm{\pi}}{4}<\mathrm{x}<\frac{\mathrm{\pi}}{4}$

Solution:

Let,

$y=\cos ^{-1}\left\{\frac{\cos x+\sin x}{\sqrt{2}}\right\}$

Now

$\mathrm{y}=\cos ^{-1}\left\{\cos \mathrm{x} \frac{1}{\sqrt{2}}+\sin \mathrm{x} \frac{1}{\sqrt{2}}\right\}$

$y=\cos ^{-1}\left\{\cos x \cos \left(\frac{\pi}{4}\right)+\sin x \sin \left(\frac{\pi}{4}\right)\right\}$

$\sin \left(\frac{\mathrm{\pi}}{4}\right)=\frac{1}{\sqrt{2}} \&$

$\cos \left(\frac{\mathrm{\pi}}{4}\right)=\frac{1}{\sqrt{2}}$

Using,

$\cos (A-B)=\cos A \cos B+\sin A \sin B$

$y=\cos ^{-1}\left\{\cos \left(x-\frac{\pi}{4}\right)\right\}$

Considering the limits,

$\frac{-\pi}{4}<\mathrm{x}<\frac{\mathrm{\pi}}{4}$

$-\frac{\pi}{4}-\frac{\pi}{4}

$-\frac{2 \pi}{4}

$\frac{-\pi}{2}

Now,

$y=\cos ^{-1}\left\{\cos \left(x-\frac{\pi}{4}\right)\right\}$

$\mathrm{y}=-\left(\mathrm{x}-\frac{\mathrm{\pi}}{4}\right)$                                                                                                                        $\left\{\begin{array}{l} \cos ^{-1}(\cos \theta)=-\theta \\ \text { if } \theta \in[-\pi, 0] \end{array}\right\}$

Differentiating with respect to x , We get

$\frac{d y}{d x}=-1$

$\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\text { constants })=0$

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