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If \( f(x)=|\cos x-\sin x| \), then \( f^{\prime}\left(\frac{\Pi}{6}\right) \) is equal to
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Verified Answer
The correct answer is:
\( -\frac{1}{2}(1+\sqrt{3}) \)
Given that, $f(x)=|\cos x-\sin x|$
So, $f^{\prime}(x)=-\sin x-\cos x$
At $x=\frac{\Pi}{6}$, we have
$f^{\prime}\left(\frac{\Pi}{6}\right)=-\sin \left(\frac{\Pi}{6}\right)-\cos \left(\frac{\Pi}{6}\right)$
$=-\frac{1}{2}-\frac{\sqrt{3}}{2}=\frac{-1}{2}(1+\sqrt{3})$
So, $f^{\prime}(x)=-\sin x-\cos x$
At $x=\frac{\Pi}{6}$, we have
$f^{\prime}\left(\frac{\Pi}{6}\right)=-\sin \left(\frac{\Pi}{6}\right)-\cos \left(\frac{\Pi}{6}\right)$
$=-\frac{1}{2}-\frac{\sqrt{3}}{2}=\frac{-1}{2}(1+\sqrt{3})$
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