Mathematics - Continuity - Differentiability Question with Solution | TestHub

Let $f(x)=x|x|=x|x|,g(x)=\sin x$ and $h(x)=gof(x)=g[f(x)]$ $$\begin{array}{rl}& \therefore h(x)=\{\begin{array}{ll}\sin x^2,& x\ge 0\\ -\sin x^2,& x<0\end{array}\\ & \text{ Now, }{h}^{\prime }(x)=\{\begin{array}{cc}2x\cos x^2,& x\ge 0\\ -2x\cos x^2,& x<0\end{array}\end{array}$$ Since, L.H.L and R.H.L at $x=0$ of $h^{\prime }(x)$ is equal to 0 therefore $h^{\prime }(x)$ is continuous at $x=0$ Now, suppose $h^{\prime }(x)$ is differentiable $$\begin{array}{rl}& \therefore h^{\prime \prime }(x)\\ & =\{\begin{array}{ll}2(\cos x^2+2x^2\left(-\sin x^2\right),& x\ge 0\\ 2\left(-\cos x^2+2x^2\sin x^2\right),& x<0\end{array}\end{array}$$ Since, L.H.L and R.H.L at $x=0$ of $h^{\prime \prime }(x)$ are different therefore $h^{\prime \prime }(x)$ is not continuous. $\Rightarrow h^{\prime \prime }(x)$ is not differentiable $\Rightarrow$ our assumption is wrong Hence $h^{\prime }(x)$ is not differentiable at $x=0$.