Let f(x)=x|x| and g(x)=sin x Statement-1: gof is differentiable at x=…

$g(f(x)) =\begin{cases}-\sin x^2, & x<0 \\ \sin x^2, & x \geq 0\end{cases}$

$LHL]_{x=0^-}=0$

$RHL]_{x=0^+}=0$

$[g(f(x))]'=\begin{cases}-2x\cos x^2, & x<0 \\ 2x\cos x^2, & x \geq 0\end{cases}$

$LHD]_{x=0^-}=0$

$RHD]_{x=0^+}=0$

$LHD=RHD→gof(x)$ differentiable at $x = 0$ and continuous derivable at $x = 0$

RHD of $(g(f(0)))'=\lim\limits_{x \rightarrow 0^{+}}\frac{2h\cos h^2-gof'(0)}{h}=2$

LHS of $(g(f(0)))'=\lim\limits_{x \rightarrow 0^{-}}\frac{-2h\cos h^2-gof'(0)}{-h}=2$

$gof$ is twice differentiable at $x = 0$