$g(x) = \left\{ {\begin{array}{*{20}{r}}{{x^2}\,\sin \frac{1}{x}\,\,,}&{x \ne 0}\\{0\,\,\,\,,}&{x = 0}\end{array}} \right.$
$L\,f'(0) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(0 - h) - f(0)}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \,\frac{{(0 - h)\,\sin \,( - \frac{1}{h}) - (0)}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \,\, - \sin \,\left( {\frac{1}{h}} \right)$
$=$ a quantity which lies between $-1$ and $1$
$R\,f'(0) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(0 + h) - f(0)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \,\frac{{(0 + h)\,\,\sin \frac{1}{h} - 0}}{h} = \mathop {\lim }\limits_{h \to 0} \,\,\sin \frac{1}{h}$
$=$ a quantity which lies between $-1$ and $1$
Hence $L\,f'(0) \ne R\,f'(0)$
$\therefore$ $f(x)$ is not differentiable at $x = 0$
now $L\,g'(0) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(0 - h) - f(0)}}{{0 - h}}$
$L\,g'(0) = \mathop {\lim }\limits_{h \to 0} \,\frac{{{{(0 - h)}^2}\sin \,( - \frac{1}{h}) - 0}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} h\,\sin \,\left( {\frac{1}{h}} \right)$
$L\,g'(0) = 0 \times \left( { - 1 \le \sin \frac{1}{h} \le 1} \right)$ ==> $L\,g'(0) = 0$
and $Rg'(0) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f\,(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{{(0 + h)}^2}\sin \left( {\frac{1}{h}} \right) - 0}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} h\,\sin \left( {\frac{1}{h}} \right) = 0 \times \left( { - 1 \le \sin \left( {\frac{1}{h}} \right) \le 1} \right) = 0$
$\because \,\,L\,{g}'(0)=R\,{g}'(0)$, then $g(x)$, is differentiable at $x=0$
now $g(x) = {x^2}\sin \frac{1}{x}$
$g'(x) = 2x\,\sin \frac{1}{x} + {x^2}\cos \frac{1}{x} \times - \frac{1}{{{x^2}}}$
$g'(x) = 2x\,\sin \frac{1}{x} - \cos \frac{1}{x}$ $ \Rightarrow g'(x) = 2\,f(x) - \cos \frac{1}{x}$
So, $g'(x)$ is not differentiable at $x = 0$.
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