Let $y = \cos^{-1}\left(\frac{\sin x + \cos x}{\sqrt{2}}\right)$. We can rewrite the argument as: $$ \frac{\sin x}{\sqrt{2}} + \frac{\cos x}{\sqrt{2}} = \cos x \cdot \frac{1}{\sqrt{2}} + \sin x \cdot \frac{1}{\sqrt{2}} $$ Using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$, we set $\cos B = \frac{1}{\sqrt{2}}$ and $\sin B = \frac{1}{\sqrt{2}}$, which implies $B = \frac{\pi}{4}$.
The expression becomes: $$ y = \cos^{-1}\left(\cos x \cos \frac{\pi}{4} + \sin x \sin \frac{\pi}{4}\right) $$ $$ y = \cos^{-1}\left(\cos\left(x - \frac{\pi}{4}\right)\right) $$
Given the domain $-\frac{\pi}{4} < x < \frac{\pi}{4}$, we have $-\frac{\pi}{2} < x - \frac{\pi}{4} < 0$. Since $\cos^{-1}(\cos \theta) = \theta$ for $\theta \in [0, \pi]$, we use the property $\cos(-\theta) = \cos \theta$: $$ y = \cos^{-1}\left(\cos\left(\frac{\pi}{4} - x\right)\right) = \frac{\pi}{4} - x $$
Now, differentiate $y$ with respect to $x$: $$ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} - x\right) = 0 - 1 = -1 $$
Final Answer: -1
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