Is f(x) = |sinx| + |cos x| differentiable at x = pi/2?

Solution 1:

For $x\to\frac{\pi}{2}^-$, $|\sin x| = \sin x$ and $|\cos x| = \cos x$

For $x\to\frac{\pi}{2}^+$, $|\sin x| = \sin x$ and $|\cos x| = -\cos x$

$\begin{align}RHD &= \lim_{x\to\frac{\pi}{2}^+}\frac{f(x) - f(\frac{\pi}{2})}{x-\frac{\pi}{2}}=\lim_{h\to0}\frac{f(\frac{\pi}{2}+h)-f(\frac{\pi}{2}) }{h}\end{align}$

Now for $x\to\frac{\pi}{2}^+$$|\sin x| = \sin x$ and $|\cos x| = -\cos x$

So,

$\begin{align}RHD &= \lim_{h\to0}\frac{\sin(\frac{\pi}{2}+h)-\cos(\frac{\pi}{2}+h)-1}{h} = \lim_{h\to0}\frac{\cos(h)+\sin(h)-1}{h}\\&=\lim_{h\to0}\frac{\cos(h)-1}{h}+\lim_{h\to0}\frac{\sin(h)}{h}=0+1 = 1\end{align} $

Similarly find, $LHD$. Check if it's equal to $RHD$