Let f(x)=x|x|, g(x)=sin x and h(x)=(g ° f)(x). Then

Let $f(x)=x|x|=x|x|, g(x)=\sin x$f(x)=x∣x∣=x∣x∣,g(x)=sinx and $h(x)=g o f(x)=g[f(x)]$h(x)=gof(x)=g[f(x)] $$\begin{aligned} &\therefore h(x)= \begin{cases}\sin x^2, & x \geq 0 \\ -\sin x^2, & x < 0\end{cases} \\ &\text { Now, } h^{\prime}(x)=\left\{\begin{array}{cc} 2 x \cos x^2, & x \geq 0 \\ -2 x \cos x^2, & x < 0 \end{array}\right. \end{aligned}$$​∴h(x)={sinx2,−sinx2,​x≥0x<0​ Now, h′(x)={2xcosx2,−2xcosx2,​x≥0x<0​​ Since, L.H.L and R.H.L at $x=0$x=0 of $h^{\prime}(x)$h′(x) is equal to 0 therefore $h^{\prime}(x)$h′(x) is continuous at $x=0$x=0 Now, suppose $h^{\prime}(x)$h′(x) is differentiable $$\begin{aligned} &\therefore h^{\prime \prime}(x) \\ &= \begin{cases}2\left(\cos x^2+2 x^2\left(-\sin x^2\right),\right. & x \geq 0 \\ 2\left(-\cos x^2+2 x^2 \sin x^2\right), & x < 0\end{cases} \end{aligned}$$​∴h′′(x)={2(cosx2+2x2(−sinx2),2(−cosx2+2x2sinx2),​x≥0x<0​​ Since, L.H.L and R.H.L at $x=0$x=0 of $h^{\prime \prime}(x)$h′′(x) are different therefore $h^{\prime \prime}(x)$h′′(x) is not continuous. $\Rightarrow h^{\prime \prime}(x)$⇒h′′(x) is not differentiable $\Rightarrow$⇒ our assumption is wrong Hence $h^{\prime}(x)$h′(x) is not differentiable at $x=0$x=0.