If f(x) = |cos x - sin x|,then f^{prime}left( | vedclass

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(A) Given the function $f(x) = |\cos x - \sin x|$.In the neighborhood of $x = \frac{\pi}{6}$,we have $\cos x > \sin x$,so $f(x) = \cos x - \sin x$.Now

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$-\frac{1}{2}(1+\sqrt{3})$

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(A) Given the function $f(x) = |\cos x - \sin x|$.In the neighborhood of $x = \frac{\pi}{6}$,we have $\cos x > \sin x$,so $f(x) = \cos x - \sin x$.Now,differentiate $f(x)$ with respect to $x$:$f^{\prime}(x) = \frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x$.Substitute $x = \frac{\pi}{6}$ into the derivative:$f^{\prime}\left(\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) - \cos\left(\frac{\pi}{6}\right)$.Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$,we get:$f^{\prime}\left(\frac{\pi}{6}\right) = -\frac{1}{2} - \frac{\sqrt{3}}{2} = -\frac{1}{2}(1 + \sqrt{3})$.

If $f(x) = |\cos x - \sin x|$,then $f^{\prime}\left(\frac{\pi}{6}\right)$ is equal to

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