Let f(x)=x∣x∣=x∣x∣,g(x)=sinx and h(x)=gof(x)=g[f(x)] ∴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 of h′(x) is equal to 0 therefore h′(x) is continuous at x=0 Now, suppose h′(x) is differentiable ∴h′′(x)={2(cosx2+2x2(−sinx2),2(−cosx2+2x2sinx2),x≥0x<0 Since, L.H.L and R.H.L at x=0 of h′′(x) are different therefore h′′(x) is not continuous. ⇒h′′(x) is not differentiable ⇒ our assumption is wrong Hence h′(x) is not differentiable at x=0.