Question:
$\cos ^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right), \frac{-\pi}{4}
Solution:
Let $y=\cos ^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right)$
$=\cos ^{-1}\left[\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right]$
$=\cos ^{-1}\left[\sin \frac{\pi}{4} \sin x+\cos \frac{\pi}{4} \cdot \cos x\right]=\cos ^{-1}\left[\cos \left(\frac{\pi}{4}-x\right)\right]$
$y=\frac{\pi}{4}-x$ $\left[\because-\frac{\pi}{4} Differentiating both sides w.r.t. $x$ Thus, $\quad \frac{d y}{d x}=-1$
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