Where is the function h(x)=|x-1| |x 2| differentiable? give a formula for h' and sketch the graphs of h and h'.

I'm guessing the second part is the absolute Val of x^2.
One important thing to note before we get started is that the only time absolute values make a function not differentiable is when it changes the function, in other words causes it to have a sharp turn.
We know that x^2 and x-2 are functions that are differentiable everywhere, so the only points of discontinuity in their derivatives will have to come from the absolute values.
anyways, x^2 is already a positive function throughout (it doesn't go below the x axis) so, you can drop the abs values for because it's always positive, so |x^2| = x^2.

x -2 on the other hand DOES indeed go below the x-axis, so it will be not differentiable at the point where it crosses the x-axis, thus the point where the function equals zero, so we solve x-2 = 0 and we get x = 2

so, the function is differentiable from negative infinity to 2 (2 not included) and 2 to infinity.

(-inf, 2) U (2, inf).