Solve this


If $\pi \leq x \leq 2 \pi$ and $y=\cos ^{-1}(\cos x)$, find $\frac{d y}{d x}$


$y=\cos ^{-1}(\cos x)$

for $x \in(\pi, 2 \pi)$

$y=\cos ^{-1}(\cos x)$

$=\cos ^{-1}(\cos (\pi+(x-\pi)))$

$=\cos ^{-1}(-\cos (x-\pi))$


$=2 \pi-x$

[Since, $\cos (\pi+x)=-\cos x$ and $\left.\cos ^{-1}(-x)=\pi-x\right]$

So, $\frac{\mathrm{dy}}{\mathrm{dx}}=-1$

For $\cos ^{-1}(\cos x), x=n_{\pi}$ are the 'sharp corners' where slope changes from 1 to $-1$ or vice versa, i.e., the points where the curves are not differentiable.

So, for $x \in[\pi, 2 \pi]$

$\frac{d y}{d x}=\left\{\begin{array}{c}-1, x \in(\pi, 2 \pi) \\ \text { does not exist for } x=\pi, 2 \pi\end{array}\right.$ (Ans)

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