Let f(x) = x|x|, g(x) = sinx and h(x) = (gof)(x). Then h(x) is not differentiable at x=0. h(x) is differentiable at x=0, but h'(x) is not continuous at x=0. h'(x) is differentiable at x=0. h'(x) is continuous at x=0 but it is not differentiable at x=0.

We have, f(x) = -x², x<0
f(x) = x², x≥0
g(x) = sinx
Hence, h(x) = g(f(x)) = sin(-x²), x<0
h(x) = g(f(x)) = sin(x²), x≥0
As x→0, h(x)→0. h(0) = 0. Hence, h(x) is continuous at x=0.
For h(x), at x=0,
L.H.D = limx→0- (h(x) - h(0))/(x-0) = limx→0- sin(-x²)/x = 0
R.H.D = limx→0+ (h(x) - h(0))/(x-0) = limx→0+ sin(x²)/x = 0
Hence, h(x) is differentiable at x=0.
We have, h'(x) = -2xcos(-x²), x<0
h'(x) = 2xcos(x²), x≥0
Let us consider the derivative of h'(x) at x=0,
L.H.D = limx→0- (h'(x) - h'(0))/(x-0) = limx→0- (-2xcos(-x²))/x = -2
R.H.D = limx→0+ (h'(x) - h'(0))/(x-0) = limx→0+ (2xcos(x²))/x = 2
Hence, h'(x) is not differentiable at x=0.