Let S=t∈ R:f(x)=|x-π | · (e^|x|-1) sin|x| is not differentiable at t. Then, the set S is equal to:

$$f(x)=|x-\pi |.({e}^{|x|}-1).\mathrm{sin}|x|$$f(x)=∣x−π∣.(e∣x∣−1).sin∣x∣

In the neighbourhood of $x=\pi$x=π, $f(x)=|x-\pi |.({e}^{x}-1).\mathrm{sin}x$f(x)=∣x−π∣.(ex−1).sinx

$$R{f}^{'}(\pi )=\underset{h\rightarrow 0}{\mathrm{lim}}⁡\frac{f(\pi +h)-f(\pi )}{h}=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{|h|\mathrm{sin}⁡(\pi +h)({e}^{\pi +h}-1)}{h}=0$$Rf′(π)=h→0lim​⁡hf(π+h)−f(π)​=h→0lim​h∣h∣sin⁡(π+h)(eπ+h−1)​=0

$$L{f}^{'}(\pi )=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(\pi )-f(\pi -h)}{h}=-\underset{h\rightarrow 0}{\mathrm{lim}}\frac{|-h|\mathrm{sin}⁡(\pi -h)({e}^{\pi -h}-1)}{h}=0$$Lf′(π)=h→0lim​hf(π)−f(π−h)​=−h→0lim​h∣−h∣sin⁡(π−h)(eπ−h−1)​=0

Differentiable at $x=\pi$x=π

$$R{f}^{'}(0)=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(h)-f(0)}{h}=\underset{h\rightarrow 0}{\mathrm{lim}}|h-\pi |\times \frac{\mathrm{sin}|h|({e}^{|h|}-1)}{h}=0$$Rf′(0)=h→0lim​hf(h)−f(0)​=h→0lim​∣h−π∣×hsin∣h∣(e∣h∣−1)​=0

$$L{f}^{'}(0)=\underset{h\rightarrow 0}{\mathrm{lim}}⁡\frac{f(0)-f(-h)}{h}=-\underset{h\rightarrow 0}{\mathrm{lim}}|-h-\pi |\times \frac{\mathrm{sin}|-h|({e}^{|-h|}-1)}{h}=0$$Lf′(0)=h→0lim​⁡hf(0)−f(−h)​=−h→0lim​∣−h−π∣×hsin∣−h∣(e∣−h∣−1)​=0

Differentiable at $x=0$x=0

Hence, $f(x)$f(x) is differentiable everywhere.

$$\Rightarrow S=\phi$$⇒S=ϕ