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Q.
If $f ( x )=|\cos x -\sin x |$, then $f '\left(\frac{\pi}{4}\right)$ is equal to
Limits and Derivatives
Solution:
We have, $f(x)=|\cos x-\sin x|$
$\Rightarrow f(x) =
\begin{cases}
\cos x-\sin x, for 0< x \leq \frac{\pi}{4} \\
\sin x-\cos x, for \frac{\pi}{4}< x < \frac{\pi}{2}
\end{cases}$
Clearly, $Lf'\left(\frac{\pi}{4}\right) =\left\{\frac{ d }{ dx }(\cos x -\sin x )\right\}_{ at\, x =\frac{\pi}{4}}$
$=(-\sin x -\cos x )_{ x -\frac{\pi}{4}}=\sqrt{2}$
and $Rf'\left(\frac{\pi}{4}\right)=\left\{\frac{ d }{ dx }(\sin x-\cos x)\right\}_{\text {at } x=\frac{\pi}{4}}$
$=(\cos x+\sin x)_{x=\frac{\pi}{4}}=\sqrt{2}$
$\because Lf'\left(\frac{\pi}{4}\right) \neq Rf'\left(\frac{\pi}{4}\right) \therefore f '\left(\frac{\pi}{4}\right)$ doesn't exist.